C H A P T E R 9
Infinite Series
Section 9.1
Sequences............................................................................................258
Section 9.2
Series and Convergence .....................................................................272
Section 9.3
The Integral Test and p-Series ...........................................................287
Section 9.4
Comparisons of Series........................................................................299
Section 9.5
Alternating Series ...............................................................................306
Section 9.6
The Ratio and Root Tests...................................................................315
Section 9.7
Taylor Polynomials and Approximations .........................................328
Section 9.8
Power Series .......................................................................................339
Section 9.9
Representation of Functions by Power Series...................................352
Section 9.10
Taylor and Maclaurin Series ..............................................................360
Review Exercises .......................................................................................................377
Problem Solving .........................................................................................................390
© 2010 Brooks/Cole, Cengage Learning
C H A P T E R
Infinite Series
9
Section 9.1 Sequences
1. an = 3n
nπ
2
5. an = sin
1
a1 = 3 = 3
π
a2 = 32 = 9
a1 = sin
a3 = 33 = 27
a2 = sin π = 0
a4 = 3 = 81
3π
= −1
2
a4 = sin 2π = 0
4
a3 = sin
a5 = 3 = 243
5
3n
n!
3
= 3
a1 =
1!
2. an =
a2 =
6. an =
32
9
=
2!
2
a1 =
a2 =
4
3
81
27
=
=
4!
24
8
35
243
81
=
=
a5 =
5!
120
40
( )
3. an = − 14
n
a3 =
a4 =
a5 =
a1 = − 14
( )
= ( − 14 )
= ( − 14 )
= ( − 14 )
a2 =
a3
a4
a5
− 14
( )
4. an = − 23
a1 = − 23
a2 =
4
9
8
a3 = − 27
a4 =
2
3
4
5
n
5π
=1
2
a5 = sin
33
27
9
=
=
a3 =
3!
6
2
a4 =
=
=1
2
1
16
1
= − 64
=
1
256
=
1
− 1024
7. an =
2n
n +3
2
1
=
4
2
4
5
6
=1
6
8
7
10
5
=
8
4
(−1)n ( n +1) 2
n2
(−1)
1
a1 =
a2 =
a3 =
12
(−1)3
22
(−1)6
32
(−1)
= −1
= −
=
1
9
=
1
16
10
a4 =
42
(−1)
1
4
15
a5 =
5
2
= −
1
25
16
81
32
a5 = − 243
258
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
8. an = ( −1)
a1 =
a2 =
a3 =
a4 =
a5 =
n +1 ⎛ 2 ⎞
⎜ ⎟
⎝n⎠
2
= 2
1
2
− = −1
2
2
3
2
1
− = −
4
2
2
5
1
1
+ 2
n
n
a1 = 5 − 1 + 1 = 5
9. an = 5 −
1
2
1
a3 = 5 −
3
1
a4 = 5 −
4
1
a5 = 5 −
5
a2 = 5 −
1
19
=
4
4
1
43
+
=
9
9
1
77
+
=
16
16
1
121
+
=
25
25
+
2
6
10. an = 10 + + 2
n
n
a1 = 10 + 2 + 6 = 18
3
25
=
2
2
2
34
+
=
3
3
3
87
+
=
8
8
6
266
+
=
25
25
a2 = 10 + 1 +
2
3
1
a4 = 10 +
2
2
a5 = 10 +
5
a3 = 10 +
11. a1 = 3, ak + 1 = 2( ak − 1)
a3 = 2( a2 − 1)
= 2( 4 − 1) = 6
a4 = 2( a3 − 1)
= 2(6 − 1) = 10
a5 = 2( a4 − 1)
= 2(10 − 1) = 18
259
⎛ k + 1⎞
12. a1 = 4, ak + 1 = ⎜
⎟ ak
⎝ 2 ⎠
⎛1 + 1⎞
a2 = ⎜
⎟ a1 = 4
⎝ 2 ⎠
⎛ 2 + 1⎞
a3 = ⎜
⎟ a2 = 6
⎝ 2 ⎠
⎛ 3 + 1⎞
a4 = ⎜
⎟ a3 = 12
⎝ 2 ⎠
⎛ 4 + 1⎞
a5 = ⎜
⎟ a4 = 30
⎝ 2 ⎠
1a
2 k
13. a1 = 32, ak + 1 =
(32)
= 16
1
2
(16)
= 8
=
1
2
(8)
= 4
=
1
2
( 4)
= 2
a2 =
1a
2 1
=
1
2
a3 =
1
a
2 2
=
a4 =
1a
2 3
a5 =
1a
2 4
14. a1 = 6, ak + 1 =
a2 =
1a 2
3 1
=
a3 =
1 2
a
3 2
=
a4 =
1a 2
3 3
=
a5 =
1a 2
3 4
=
15. an =
1a 2
3 k
(62 ) = 12
1
122 ) = 48
3(
1 482 = 768
)
3(
1
3
1
3
(768)
2
= 196,608
10
10
10
, a1 =
= 5, a2 =
1+1
3
n +1
Matches (c)
16. an =
10n
10
20
, a1 =
= 5, a2 =
n +1
2
3
Matches (a)
17. an = ( −1) , a1 = −1, a2 = 1, a3 = −1, …
n
Matches (d)
a2 = 2( a1 − 1)
= 2(3 − 1) = 4
Sequences
18. an =
(−1)
n
n
, a1 =
−1
1
= −1, a2 = .
1
2
Matches (b)
2
19. −2, 0, , 1, … matches (b)
3
4
n
4
4
4
2
a1 = 2 − = −2, a2 = 2 − = 0, a3 = 2 − = , …
1
2
3
3
an = 2 −
© 2010 Brooks/Cole, Cengage Learning
260
Chapter 9
Infinite Series
20. 16, − 8, 4, − 2, … matches (c)
an = 16( −0.5)
n −1
a1 = 16( −0.5)
1 −1
21.
2, 4,
3 3
28. an = − 32 an −1 , a1 = 1
= 16, a2 = 16(−0.5)
2 −1
2, 83 , … matches (a)
an =
2n
3
a1 =
2,
3
a2 =
4,
3
a3 =
6
3
= −8, …
29.
= 2, …
2n
n +1
2
4
6
3
= 1, a2 = , a3 =
= ,…
a1 =
2
3
4
2
an =
30.
31.
32.
23. an = 3n − 1
a5 = 3(5) − 1 = 14
n + 6
24. an =
2
5+ 6
11
=
a5 =
2
2
6+ 6
= 6
a6 =
2
1
Add to preceding term.
2
25. an + 1 = 2an , a1 = 5
11! 11(10)(9)8!
=
= 11(10)(9) = 990
8!
8!
25( 24)( 23)( 22)( 21)( 20)!
25!
=
20!
20!
= 25( 24)( 23)( 22)( 21) = 6,375,600
(n
+ 1)!
n!
(n
(2n − 1)!
(2n + 1)!
=
34.
(2n + 2)!
(2n)!
=
1,
16
1
a6 = − 32
Multiply the preceding term by − 12 .
27. an =
a5 =
a6 =
3
(−2)
35. lim
n→∞
3
(−2)
4
3
(−2)
5
=
3
16
= −
3
32
1
Multiply the preceding term by − .
2
(2n − 1)!
1
=
− 1)!( 2n)( 2n + 1)
2n( 2n + 1)
(2n)!(2n
+ 1)( 2n + 2)
(2n)!
+ 1)( 2n + 2)
5n 2
= 5
n + 2
2
1⎞
⎛
36. lim ⎜ 5 − 2 ⎟ = 5 − 0 = 5
n→∞ ⎝
n ⎠
37. lim
n→∞
38. lim
n→∞
2n
n2 + 1
5n
n2 + 4
2
= lim
n→∞
= lim
n→∞
=
1 + (1 n 2 )
5
1 + (4 n
2
)
=
2
= 2
1
5
= 5
1
⎛1⎞
39. lim sin ⎜ ⎟ = 0
n→∞
⎝n⎠
⎛ 2⎞
40. lim cos⎜ ⎟ = 1
n→∞
⎝n⎠
41.
n −1
( 2n
= ( 2n
Multiply the preceding term by 2.
26. an = − 12 an −1 , a1 = 1
= n +1
n!
+ 2)!
n!( n + 1)( n + 2)
=
= ( n + 1)( n + 2)
n!
n!
a5 = 2( 40) = 80
a6 = 2(80) = 160
n!( n + 1)
=
33.
a6 = 3(6) − 1 = 17
Add 3 to preceding term.
a6 = − 243
32
Multiply the preceding term by − 32 .
4 3 8
22. 1, , , , … matches (d)
3 2 5
a5 =
81
,
16
a5 =
3
−1
12
−1
The graph seems to indicate that the sequence converges
to 1. Analytically,
lim an = lim
n→∞
n→∞
n +1
x +1
= lim
= lim 1 = 1.
x→∞
x→∞
n
x
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
42.
2
51. lim
n→∞
−1
n
= 1, converges
n +1
n→∞ 3
1
1
lim an = lim 3 2 = lim 3 2 = 0.
n→∞
n→∞ n
x→∞ x
43.
3
52. lim
The graph seems to indicate that the sequence converges
to 0. Analytically,
261
3n 2 − n + 4
3
= , converges
2n 2 + 1
2
12
−1
Sequences
53. an =
=
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1)
( 2n ) n
1
3
5
2n − 1
1
⋅
⋅
⋅⋅⋅
<
2n 2n 2n
2n
2n
So, lim an = 0, converges.
2
n→∞
−1
12
−2
The graph seems to indicate that the sequence diverges.
Analytically, the sequence is
{an}
= {0, −1, 0, 1, 0, −1, …}.
So, lim an does not exist.
n→∞
44.
54. The sequence diverges. To prove this analytically, you
use mathematical induction to show that
⎛ 3⎞
an ≥ ⎜ ⎟
⎝ 2⎠
n −1
⎛ 3⎞
ak ≥ ⎜ ⎟
⎝ 2⎠
k −1
. Then,
1⋅3⋅5
ak +1 =
4
0
⎛ 3⎞
. Clearly, a1 = 1 ≥ ⎜ ⎟ = 1. Assume that
⎝ 2⎠
(2k − 1)(2k
(k + 1)!
2k + 1 ⎛ 3 ⎞
≥ ⎜ ⎟
k +1
⎝ 2⎠
= ak
k −1
+ 1)
⎛ 2k + 1 ⎞
⎜
⎟.
⎝ k +1⎠
Because
−1
12
−1
The graph seems to indicate that the sequence converges
to 3. Analytically,
1⎞
⎛
lim an = lim ⎜ 3 − n ⎟ = 3 − 0 = 3.
n→∞ ⎝
2 ⎠
n→∞
n
45. lim ⎡(0.3) − 1⎤ = 0 − 1 = −1, converges
⎦
n→∞ ⎣
3⎤
⎡
46. lim ⎢4 − ⎥ = 4 − 0 = 4, converges
n→∞ ⎣
n⎦
5
47. lim
= 0, converges
n→∞ n + 2
48. lim
n→∞
2
= 0, converges
n!
n ⎞
n⎛
49. lim ( −1) ⎜
⎟
n→∞
⎝ n + 1⎠
does not exist (oscillates between −1 and 1), diverges.
2k + 1
3
≥ ,
k +1
2
k
⎛3⎞
you have ak + 1 ≥ ⎜ ⎟ , which shows that
⎝2⎠
⎛ 3⎞
an ≥ ⎜ ⎟
⎝ 2⎠
55. lim
n→∞
does not exist, (alternates between 0 and 2), diverges.
= 0, converges
1 + ( −1)
n
2
ln ( n3 )
2n
n
= 0, converges
3 ln ( n)
n→∞ 2
n
3⎛1⎞
= lim ⎜ ⎟ = 0, converges
n→∞ 2 ⎝ n ⎠
= lim
(L'Hôpital's Rule)
58. lim
n→∞
n
50. lim ⎡1 + ( −1) ⎤
⎦
n→∞ ⎣
n
n
n→∞
57. lim
for all n.
1 + ( −1)
n→∞
56. lim
n −1
ln
n
n
= lim
(1 2) ln n
n
1
= lim
= 0, converges
n → ∞ 2n
n→∞
(L'Hôpital's Rule)
© 2010 Brooks/Cole, Cengage Learning
262
Chapter 9
59. lim
n→∞
( 34 )
n
Infinite Series
1n
n
n2
⎡
1⎞
1 ⎞ ⎤⎥
⎛
⎛
⎢
70. an = ⎜1 + 2 ⎟ = ⎜1 + 2 ⎟
⎢⎝
n ⎠
n ⎠ ⎥
⎝
⎣
⎦
= 0, converges
60. lim (0.5) = 0, converges
n
lim an = e1 n = 1, converges
n→∞
n→∞
(n
61. lim
+ 1)!
= lim ( n + 1) = ∞, diverges
n →∞
n!
n →∞
(n
62. lim
− 2)!
= lim
n!
n→∞
n→∞
1
= 0, converges
n( n − 1)
(n − 1) − n2
2
n ⎞
⎛n − 1
63. lim ⎜
−
⎟ = nlim
n→∞ ⎝
→∞
n
n − 1⎠
n( n − 1)
1 − 2n
= lim 2
= 0, converges
n→∞ n − n
⎛ n2
n2 ⎞
1
−2n 2
64. lim ⎜
−
= lim 2
= − , converges
⎟
n →∞ 2 n + 1
2n − 1 ⎠ n →∞ 4n − 1
2
⎝
n→∞
(p
n
= 0, converges
en
> 0, n ≥ 2)
66. an = n sin
n→∞
sin n
1
= lim (sin n) = 0,
n
→
∞
n
n
converges (because (sin n) is bounded)
72. lim
n→∞
cos π n
= 0, converges
n2
73. an = 3n − 2
74. an = 4n − 1
75. an = n 2 − 2
76. an =
p
65. lim
71. lim
77. an =
1
n
78. an =
1
Let f ( x) = x sin .
x
(−1 x ) cos(1 x)
2
−1 x 2
x →∞
n
1
= cos 0
x
= 1 (L'Hôpital's Rule)
= lim cos
n −1
2
n +1
n + 2
(−1)
2
n −1
n−2
79. an = 1 +
sin (1 x)
1
lim x sin = lim
x →∞
x
→∞
x
1x
= lim
(−1)
1
n +1
=
n
n
2n − 1
2n
n +1
−1
2
=
2n
80. an = 1 +
x →∞
81. an =
or,
lim
x→∞
sin (1 x)
lim n sin
n→∞
= lim
1x
y → 0+
sin ( y )
y
= 1. Therefore
1
= 1, converges.
n
1
n!
83. an =
(−1)n −1
1 ⋅ 3 ⋅ 5 ( 2n
n→∞
n→∞
n→∞
k⎞
⎛
69. an = ⎜1 + ⎟
n⎠
⎝
−1
= 0, converges
3n
n
n
k⎞
1u k
⎛
lim ⎜1 + ⎟ = lim ⎡(1 + u ) ⎤ = e k
⎣
⎦
→
0
n→∞ ⎝
u
n⎠
where u = k n, converges
+ 1)( n + 2)
82. an =
67. lim 21 n = 20 = 1, converges
68. lim − 3− n = lim
(n
n
=
84. an =
− 1)
(−1)n −1 2n n!
(2n)!
x n −1
(n − 1)!
85. an = ( 2n)!, n = 1, 2, 3, …
86. an = ( 2n − 1)!, n = 1, 2, 3, …
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
1
1
< 4−
= an + 1 ,
n
n +1
87. an = 4 −
Monotonic; an < 4, bounded
88. Let f ( x) =
6
3x
.
. Then f ′( x) =
x + 2
( x + 2) 2
So, f is increasing which implies {an} is increasing.
n
2
n +1
?
n+2
≥
2(n + 1) + 2
?
n ≥1
2n ≥ n + 1
2n + 3 n ≥ 2n + 2 ( n + 1)
n +1
≥ (n + 1) + 2
2n + 2
2
an ≥ an + 1.
1
, bounded
8
( 32 )
n
<
( 32 )
n +1
bounded
= an + 1
Monotonic; lim an = ∞, not bounded
n→∞
⎛ nπ ⎞
96. an = cos⎜ ⎟
⎝ 2 ⎠
a1 = cos
π
2
= 0
⎛ 3π ⎞
a3 = cos⎜ ⎟ = 0
⎝ 2 ⎠
Not monotonic; an ≤ 1, bounded
cos n
n
a1 = 0.5403
a1 = 0.6065
a2 = −0.2081
a3 = 0.6694
Not monotonic; an ≤ 0.7358, bounded
a3 = −0.3230
a4 = −0.1634
Not monotonic; an ≤ 1, bounded
n⎛ 1 ⎞
91. an = ( −1) ⎜ ⎟
⎝n⎠
a1 = −1
98. an =
1
2
a1 =
1
a3 = −
3
Not monotonic; an ≤ 1, bounded
a2 =
a3 =
n
sin n
n
sin (1)
1
sin ( 2)
2
sin (3)
3
≈ 0.8415
a4 =
≈ 0.4546
a5 =
≈ 0.0470
a6 =
sin ( 4)
4
sin (5)
5
sin (6)
6
≈ −0.1892
≈ −0.1918
≈ −0.0466
Not monotonic; an ≤ 1, bounded
a1 = − 23
a2 =
94. an =
2,
3
97. an =
a2 = 0.7358
( )
Monotonic; an ≤
= an + 1
a2 = cos π = −1
90. an = ne−n 2
92. an = − 23
n +1
Not monotonic; an ≤ 1, bounded
n ≥ 1
a2 =
( 23 )
a4 = 0.8660
2n ≥ n + 1
Monotonic; an ≤
>
a3 = 1.000
2n + 3 n ≥ 2n + 2 ( n + 1)
n
n
a2 = 0.8660
?
So,
( 23 )
263
⎛ nπ ⎞
95. an = sin ⎜ ⎟
⎝ 6 ⎠
a1 = 0.500
an < 3, bounded
89.
93. an =
Sequences
4
9
8
a3 = − 27
Not monotonic; an ≤
2
, bounded
3
© 2010 Brooks/Cole, Cengage Learning
264
Chapter 9
Infinite Series
99. (a) an = 5 +
5+
1
n
102. (a) an = 4 +
1
≤ 6 ⇒ {an}, bounded
n
4+
1
1
> 5+
n
n +1
= an + 1 ⇒ {an}, monotonic
an = 5 +
(b)
= an + 1
−1
12
−1
1⎞
⎛
lim ⎜ 5 + ⎟ = 5
n⎠
⎝
100. (a) an = 4 −
an = 4 −
103. {an} has a limit because it is a bounded, monotonic
3
n
3
4
< 3−
= an +1 ⇒ {an}, monotonic
n
n +1
8
12
3⎞
⎛
lim ⎜ 4 − ⎟ = 4
n
⎝
⎠
2 ≤ lim an ≤ 4
n→∞
104. The sequence {an} could converge or diverge. If {an} is
increasing, then it converges to a limit less than or equal
to 1. If {an} is decreasing, then it could converge
an = 1 n) or diverge (example: an = − n).
r ⎞
⎛
105. An = P⎜1 +
⎟
12 ⎠
⎝
n
r ⎞
⎛
(a) Because P > 0 and ⎜1 +
⎟ > 1, the sequence
12
⎝
⎠
diverges. lim An = ∞
n→∞
n→∞
1⎛
1⎞
⎜1 − n ⎟
3⎝
3 ⎠
1⎛
1⎞
1
⇒ {an}, bounded
⎜1 − n ⎟ <
3⎝
3 ⎠
3
1⎛
1⎞
1⎛
1 ⎞
⎜1 − n ⎟ < ⎜1 − n + 1 ⎟
3⎝
3 ⎠
3⎝
3 ⎠
= an + 1 ⇒ {an}, monotonic
Therefore, {an} converges.
(b)
sequence. The limit is less than or equal to 4, and greater
than or equal to 2.
(example:
0
an =
12
1⎞
⎛
lim ⎜ 4 + n ⎟ = 4
2 ⎠
⎝
Therefore, {an} converges.
101. (a) an =
−1
n→∞
3
< 4 ⇒ {an}, bounded
4−
n
−1
6
−1
n→∞
(b)
1
1
> 4 + n +1
2n
2
⇒ {an}, monotonic
Therefore, {an} converges.
(b)
7
1
≤ 4.5 ⇒ {an}, bounded
2n
an = 4 +
Therefore, {an} converges.
1
2n
0.4
0.055 ⎞
⎛
(b) P = 10,000, r = 0.055, An = 10,000⎜1 +
⎟
12 ⎠
⎝
A0 = 10,000
A1 = 10,045.83
A2 = 10,091.88
A3 = 10,138.13
A4 = 10,184.60
A5 = 10,231.28
A6 = 10,278.17
A7 = 10,325.28
−1
12
−0.1
⎡1 ⎛
1 ⎞⎤
1
lim ⎢ ⎜1 − n ⎟⎥ =
3 ⎠⎦
3
⎣3⎝
A8 = 10,372.60
A9 = 10,420.14
A10 = 10,467.90
n→∞
© 2010 Brooks/Cole, Cengage Learning
n
Section 9.1
106. (a) An = 100( 401)(1.0025n − 1)
Sequences
265
113. (a) an = −5.364n 2 + 608.04n + 4998.3
A0 = 0
10,000
A1 = 100.25
A2 = 200.75
A3 = 301.50
0
4000
A4 = 402.51
8
(b) For 2012, n = 12: a12 ≈ $11,522.4 billion
A5 = 503.76
A6 = 605.27
114. (a) an = 1127.4n + 17684
(b) A60 = 6480.83
50,000
(c) A240 = 32,912.28
107. No, it is not possible. See the "Definition of the Limit of
a sequence". The number L is unique.
5
17
0
108. (a) A sequence is a function whose domain is the set of
positive integers.
(b) A sequence converges if it has a limit. See the
definition.
(b) For 2012, n = 22: a22 ≈ $42,487
10n
n!
115. an =
(c) A sequence is monotonic if its terms are
nondecreasing, or nonincreasing.
(a) a9 = a10 =
(d) A sequence is bounded if it is bounded below
(an ≥ N for some N ) and bounded above
(b) Decreasing
(an
≤ M for some M ).
109. The graph on the left represents a sequence with
alternating signs because the terms alternate from
being above the x-axis to being below the x-axis.
1
110. (a) an = 10 −
n
(c) Factorials increase more rapidly than exponentials.
1⎞
⎛
116. an = ⎜1 + ⎟
n⎠
⎝
a1 = 2.0000
3n
4n + 1
a2 = 2.2500
a4 ≈ 2.4414
a5 ≈ 2.4883
a6 ≈ 2.5216
n
(d) Impossible. An unbounded sequence diverges.
111. (a) An = (0.8) 4,500,000,000
n
(b) A1 = $3,600,000,000
1⎞
⎛
lim ⎜1 + ⎟ = e
n→∞ ⎝
n⎠
117. an =
a2 =
A3 = $2,304,000,000
A4 = $1,843,200,000
(c) lim An = lim (0.8) ( 4.5) = 0, converges
n
n→∞
112. Pn = 25,000(1.045)
P1 = $26,125.00
P2 = $27,300.63
P3 = $28,529.15
P4 = $29,812.97
P5 = $31,154.55
n
n
n = n1 n
a1 = 11 1 = 1
A2 = $2,880,000,000
n→∞
n
a3 ≈ 2.3704
(b) Impossible. The sequence converges by Theorem
9.5.
(c) an =
109
1,000,000,000
1,562,500
=
=
9!
362,880
567
2 ≈ 1.4142
a3 =
3
3 ≈ 1.4422
a4 =
4
4 ≈ 1.4142
a5 =
5
5 ≈ 1.3797
a6 =
6
6 ≈ 1.3480
Let y = lim n1 n .
n→∞
ln n
1n
⎛1
⎞
ln y = lim ⎜ ln n ⎟ = lim
= lim
= 0
n→∞ ⎝ n
n→∞ n
n→∞ n
⎠
Because ln y = 0, you have y = e0 = 1. Therefore,
lim
n→∞
n
n = 1.
© 2010 Brooks/Cole, Cengage Learning
266
Chapter 9
Infinite Series
125. an + 2 = an + an + 1
118. Because
lim sn = L > 0,
(a) a1 = 1
n→∞
a2 = 1
there exists for each ε > 0,
an integer N such that sn − L < ε for every n > N .
Let ε = L > 0 and you have,
sn − L < L, − L < sn − L < L, or 0 < sn < 2 L for
each n > N .
119. True
a9 = 21 + 13 = 34
a4 = 2 + 1 = 3
a10 = 34 + 21 = 55
a5 = 3 + 2 = 5
a11 = 55 + 34 = 89
a6 = 5 + 3 = 8
a12 = 89 + 55 = 144
120. True
b1 =
121. True
b2 =
122. True
b3 =
123. True
124. False. Let an = ( −1) and bn = ( −1)
n
converges to 0.
+ bn} =
{(−1)
n +1
n
then {an} and
+ ( −1)
}
n +1
a8 = 13 + 8 = 21
a3 = 1 + 1 = 2
(b) bn =
{bn} diverge. But {an
a7 = 8 + 5 = 13
b4 =
b5 =
(c) 1 +
an + 1
,n ≥ 1
an
1
1
2
1
3
2
5
3
8
5
=1
b6 =
= 2
b7 =
= 1.5
b8 =
≈ 1.6667
b9 =
= 1.6
b10 =
13
8
21
13
34
21
55
34
89
55
= 1.625
≈ 1.6154
≈ 1.6190
≈ 1.6176
≈ 1.6182
1
1
=1+
bn −1
an an − 1
=1+
an − 1
a + an −1
a
= n
= n +1 = bn
an
an
an
⎛
1 ⎞
(d) If lim bn = ρ , then lim ⎜1 +
⎟ = ρ.
n→∞
n→∞
b
n −1 ⎠
⎝
Because lim bn = lim bn −1 , you have
n→∞
n→∞
1 + (1 ρ ) = ρ .
ρ + 1 = ρ2
0 = ρ2 − ρ − 1
ρ =
1± 1+ 4
1± 5
=
2
2
Because an , and therefore bn , is positive,
ρ =
1+
5
2
126. x0 = 1, xn =
≈ 1.6180.
1
1
, n = 1, 2, …
xn −1 +
2
xn −1
x1 = 1.5
x6 = 1.414214
x2 = 1.41667
x7 = 1.414214
x3 = 1.414216
x8 = 1.414114
x4 = 1.414214
x9 = 1.414214
x5 = 1.414214
x10 = 1.414214
The limit of the sequence appears to be 2. In fact, this
sequence is Newton's Method applied to
f ( x) = x 2 − 2.
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
127. (a) a1 =
267
2 ≈ 1.4142
a2 =
2+
2 ≈ 1.8478
a3 =
2+
2+
2 ≈ 1.9616
a4 =
2+
2+
2+
2 ≈ 1.9904
a5 =
2+
2+
2+
2+
(b) an =
Sequences
2 + an −1 ,
n ≥ 2, a1 =
2 ≈ 1.9976
2
(c) First use mathematical induction to show that an ≤ 2; clearly a1 ≤ 2. So assume ak ≤ 2. Then
ak + 2 ≤ 4
ak + 2 ≤ 2
ak + 1 ≤ 2.
Now show that {an} is an increasing sequence. Because an ≥ 0 and an ≤ 2,
( an
− 2)( an + 1) ≤ 0
an2 − an − 2 ≤ 0
an2 ≤ an + 2
an ≤
an + 2
an ≤ an + 1 .
Because {an} is a bounding increasing sequence, it converges to some number L, by Theorem 9.5.
lim an = L ⇒
n→∞
2 + L = L ⇒ 2 + L = L2 ⇒ L2 − L − 2 = 0
⇒ ( L − 2)( L + 1) = 0 ⇒ L = 2
128. (a) a1 =
≠ −1)
6 ≈ 2.4495
a2 =
6+
6 ≈ 2.9068
a3 =
6+
6+
6 ≈ 2.9844
a4 =
6+
6+
6+
6 ≈ 2.9974
a5 =
6+
6+
6+
6+
(b) an =
(L
6 + an −1 ,
n ≥ 2, a1 =
6 ≈ 2.9996
6
(c) First use mathematical induction to show that an ≤ 3; clearly a1 ≤ 3. So assume ak ≤ 3. Then
6 + ak ≤ 9
6 + ak ≤ 3
ak + 1 ≤ 3.
Now show that {an} is an increasing sequence. Because an ≥ 0 and an ≤ 3,
(an
− 3)( an + 2) ≤ 0
an2 − an − 6 ≤ 0
an2 ≤ an + 6
an ≤
an + 6
an ≤ an + 1.
Because {an} is a bounded increasing sequence, it converges to some number L : lim an = L. So,
n→∞
6 + L = L ⇒ 6 + L = L2 ⇒ L2 − L − 6 = 0
⇒
(L
− 3)( L + 2) = 0 ⇒ L = 3
(L
≠ − 2)
© 2010 Brooks/Cole, Cengage Learning
268
Chapter 9
Infinite Series
129. (a) Use mathematical induction to show that
1+
an ≤
1 + 4k
.
2
[Note that if k = 2, and an ≤ 3, and if k = 6, then an ≤ 3. ] Clearly,
a1 =
k ≤
1 + 4k
1+
≤
2
1 + 4k
.
2
Before proceeding to the induction step, note that
2 + 2 1 + 4k + 4k = 2 + 2 1 + 4k + 4 k
1+
1 + 4k
1 + 2 1 + 4 k + 1 + 4k
+ k =
2
4
1+
⎡1 +
1 + 4k
+ k = ⎢
2
⎢⎣
1+
1 + 4k
1+
+ k =
2
So assume an ≤
an + k ≤
an + k ≤
an + 1 ≤
1+
1+
2
1 + 4k
.
2
1 + 4k
. Then
2
1 + 4k
+ k
2
1+
1+
1 + 4k ⎤
⎥
2
⎦⎥
1 + 4k
+ k
2
1 + 4k
.
2
{an} is increasing because
⎛
1+
⎜⎜ an −
⎝
1 + 4k ⎞⎛
1−
⎟⎜
⎟⎜ an −
2
⎠⎝
1 + 4k ⎞
⎟⎟ ≤ 0
2
⎠
an2 − an − k ≤ 0
an2 ≤ an + k
an ≤
an + k
an ≤ an + 1.
(b) Because {an} is bounded and increasing, it has a limit L.
(c) lim an = L implies that
n→∞
L =
k + L ⇒ L2 = k + L
⇒ L2 − L − k = 0
⇒ L =
Because L > 0, L =
1 ± 1 + 4k
.
2
1+
1 + 4k
.
2
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
Sequences
269
130. (a) a0 = 10, b0 = 3
a1 =
a2 =
a3 =
a4 =
a5 =
a0 + b0
2
a1 + b1
2
a2 + b2
2
a3 + b3
2
a4 + b4
2
=
10 + 3
= 6.5
2
10(3) ≈ 5.4772
b1 =
a0b0 =
≈ 5.9886
b2 =
a1b1 ≈ 5.9667
≈ 5.9777
b3 =
a2b2 ≈ 5.9777
≈ 5.9777
b4 =
a3b3 ≈ 5.9777
≈ 5.9777
b5 =
a4b4 ≈ 5.9777
The terms of {an} are decreasing and those of {bn} are increasing. They both seem to approach the same limit.
(b) For n = 0, you need to show a0 > a1 > b1 > b0 . Because a0 > b0 , 2a0 > a0 + b0 ⇒ a0 >
a0 + b0
= a1.
2
Because ( a0 − b0 ) > 0,
2
a02 − 2a0b0 + b02 > 0 ⇒ a02 + 2a0b0 + b02 > 4a0b0
⇒
(a0
+ b0 ) > 4a0b0 ⇒ a0 + b0 > 2 a0b0 ⇒ a1 > b1.
2
Because a0 > b0 , a0b0 > b02 ⇒
a0b0 > b0 ⇒ b1 > b0 . So, you have shown that a0 > a1 > b1 > b0 .
Now assume ak > ak + 1 > bk + 1 > bk . Because ak + 1 > bk + 1 ,
2ak + 1 > ak + 1 + bk + 1 ⇒ ak + 1 >
ak + 1 + bk + 1
= ak + 2 .
2
Because ( ak + 1 − bk + 1 ) > 0,
2
ak + 12 − 2ak + 1bk + 1 + bk + 12 > 0 ⇒ ak + 12 + 2ak + 1bk + 1 + bk + 12 > 4ak + 1bk + 1
⇒ ( ak + 1 + bk + 1 ) > 4ak + 1bk + 1
2
⇒ ak + 1 + bk + 1 > 2 ak + 1bk + 1
⇒ ak + 2 > bk + 2 .
Because ak + 1 > bk + 1 , ak + 1bk + 1 > bk + 12 ⇒
ak + 1bk + 1 > bk + 1 ⇒ bk + 2 > bk + 1.
So, you have shown that ak + 1 > ak + 2 > bk + 2 > bk + 1.
(c)
{an} converges because it is decreasing and bounded below by 0. {bn} converges because it is increasing and bounded
above by a0 .
(d) Let lim an = A and lim bn = B. Then A =
n→∞
n→∞
131. (a) f ( x) = sin x, an = n sin
A+ B
⇒ A = B.
2
1
n
f ′( x) = cos x, f ′(0) = 1
lim an = lim n sin
n→∞
n→∞
(b) f ′(0) = lim
h → 0+
sin (1 n)
1
= lim
= 1 = f ′(0)
→
∞
n
n
(1 n)
f ( 0 + h ) − f ( 0)
f ( h)
f (1 n)
⎛1⎞
= lim
= lim
= lim nf ⎜ ⎟ = lim an
+
n
n
n →∞
→∞
→∞
h
h
h→0
(1 n)
⎝n⎠
© 2010 Brooks/Cole, Cengage Learning
270
Chapter 9
Infinite Series
132. an = nr n
(c)
n
= x ln x − x + C
n
∫ 1 ln x dx
1
n
⎛1⎞
: an = n ⎜ ⎟ = n → 0
2
2
⎝ 2⎠
(a) r =
∫ ln x dx
= n ln n − n + 1 = ln n n − n + 1
From part (a): ln n n − n + 1 < ln ( n!)
(b) r = 1: an = n, diverges
eln n
2
3
⎛ 3⎞
(c) r = : an = n⎜ ⎟ , diverges
2
⎝ 2⎠
n
< n!
e n −1
nr
=
r
n +1
∫1
"∞"
∞
n
−n
ln x dx = (n + 1) ln ( n + 1) − ( n + 1) + 1
= ln ( n + 1)
1
−r n
→
=
→ 0.
−n
− r ln ( r )
ln ( r )
eln(n + 1)
(n
y
1.5
(d)
1.0
(n + 1)
nn
< n! <
en −1
en
n
0.5
2
3
4
...
<
e1 − (1 n)
x
n
∫ 1 ln x dx
< ln 2 + ln 3 +
+ ln n
= ln (1 ⋅ 2 ⋅ 3
n) = ln ( n!)
2.0
n +1 − n
+ 1)
> n!
n +1
> n!
lim
+ 1)
( n + 1)
e
1 − (1 n)
e
=
1
e
1 + (1 n)
y = lnx
lim
(n + 1)
n→∞
1.5
ne
0.5
3
4
...
= lim
(n
+ 1) ( n + 1)
1n
n
n→∞
e
1
= (1)
e
1
=
e
1.0
2
n
1 + (1 n)
1
n→∞
(n
n +1
(n + 1)
n!
<
n
ne
n
<
e1 − (1 n)
y
n! <
n
1
n
(b)
−n
en
y = lnx
2.0
n +1
From part (b): ln(n + 1) n + 1 − n > ln ( n!)
The sequence converges for r < 1.
133. (a)
< n!
n
(d) If r ≥ 1, the sequences diverges. If r < 1,
n
n − n +1
x
n+1
n
n +1
∫1
ln x dx > ln 2 + ln 3 +
+ ln n = ln ( n!)
By the Squeeze Theorem, lim
n→∞
(e) n = 20:
n = 50:
n = 100:
n!
1
= .
n
e
20
20!
≈ 0.4152
20
50
50!
≈ 0.3897
50
100
100!
≈ 0.3799
100
1
≈ 0.3679
e
© 2010 Brooks/Cole, Cengage Learning
Section 9.1
134. an =
1
1
=
= 0.5
1+1
2
a2 =
1⎡
1
1 ⎤
1 ⎡2
1⎤
7
+
≈ 0.5833
⎢
⎥ = ⎢ + ⎥ =
2 ⎣⎢1 + (1 2) 1 + 1⎦⎥
2 ⎣3
2⎦
12
a3 =
1⎡
1
1
1 ⎤
37
+
+
≈ 0.6167
⎢
⎥ =
3 ⎣⎢1 + (1 3) 1 + ( 2 3) 1 + 1⎦⎥
60
a4 =
1⎡
1
1
1
1 ⎤
533
+
+
+
≈ 0.6345
⎢
⎥ =
4 ⎣⎢1 + (1 4) 1 + ( 2 4) 1 + (3 4) 1 + 1⎦⎥
840
a5 =
1⎡
1
1
1
1
1 ⎤
1627
≈ 0.6456
+
+
+
+
⎢
⎥ =
5 ⎣⎢1 + (1 5) 1 + ( 2 5) 1 + (3 5) 1 + ( 4 5) 1 + 1⎦⎥
2520
(b) lim an = lim
n→∞
n→∞
1 n
1
=
∑
n k =1 1 + ( k n)
1
1
∫0 1 +
1
x
dx = ⎣⎡ln 1 + x ⎦⎤ 0 = ln 2
135. For a given ε > 0, you must find M > 0 such that
an − L =
1
< ε
n3
whenever n > M . That is,
13
⎛1⎞
or n > ⎜ ⎟ .
ε
⎝ε ⎠
1
So, let ε > 0 be given. Let M be an integer satisfying
M > (1 ε ) . For n > M , you have
13
13
⎛1⎞
n > ⎜ ⎟
⎝ε ⎠
1
n3 >
ε
ε >
271
1 n
1
∑
n k =1 1 + ( k n)
(a) a1 =
n3 >
Sequences
1
1
⇒ 3 − 0 < ε.
3
n
n
So, lim
n→∞
1
= 0.
n3
136. For a given ε > 0, you must find M > 0 such that
an − L = r n ε whenever n > M . That is,
n ln r < ln (ε ) or
ln (ε )
n >
ln r
(because ln r < 0 for r < 1 ).
So, let ε > 0 be given. Let M be an integer satisfying
M >
ln (ε )
ln r
.
For n > M , you have
n >
ln (ε )
ln r
n ln r < ln (ε )
ln r
n
< ln (ε )
r
n
< ε
r − 0 < ε.
n
So, lim r n = 0 for −1 < r < 1.
n→∞
137. Answers will vary. Sample answer:
{(−1) } = {−1, 1, −1, 1, …} diverges
} = {( −1) } = {1, 1, 1, 1, …} converges
{an}
{a2 n
=
n
2n
138. Let f ( x) = sin (π x)
lim sin (π x) does not exist.
x→∞
an = f ( n) = sin (π n) = 0 for all n
lim an = 0, converges
n→∞
© 2010 Brooks/Cole, Cengage Learning
272
Chapter 9
Infinite Series
139. If {an} is bounded, monotonic and nonincreasing, then a1 ≥ a2 ≥ a3 ≥
− a1 ≤ − a2 ≤ − a3 ≤
≤ − an ≤
≥ an ≥
. Then
is a bounded, monotonic, nondecreasing sequence which converges by the first half of
the theorem. Because {− an} converges, then so does {an}.
xn + 1 + xn −1
,
xn
140. Define an =
n ≥ 1.
xn + 12 − xn xn + 2 = 1 = xn2 − xn −1xn + 1 ⇒
xn + 1 ( xn + 1 + xn −1 ) = xn ( xn + xn + 2 )
+ xn
xn + 1 + xn −1
x
= n+2
xn
xn + 1
an = an + 1
Therefore, a1 = a2 = … = a. So,
xn +1 = an xn − xn −1 = axn − xn −1.
141. Tn = n! + 2n
Use mathematical induction to verify the formula.
T0 = 1 + 1 = 2
T1 = 1 + 2 = 3
T2 = 2 + 4 = 6
Assume Tk = k! + 2k . Then
Tk +1 = ( k + 1 + 4)Tk − 4( k + 1)Tk −1 + ( 4( k + 1) − 8)Tk − 2
= ( k + 5) ⎡⎣k! + 2k ⎤⎦ − 4( k + 1)(( k − 1)! + 2k −1 ) + ( 4k − 4)(( k − 2)! + 2k − 2 )
= ⎡⎣( k + 5)( k )( k − 1) − 4( k + 1)( k − 1) + 4( k − 1)⎤⎦ ( k − 2)! + ⎡⎣( k + 5)4 − 8( k + 1) + 4( k − 1)⎤⎦ 2k − 2
= ⎡⎣k 2 + 5k − 4k − 4 + 4⎤⎦ ( k − 1)! + 8 ⋅ 2k − 2
= ( k + 1)! + 2k + 1.
By mathematical induction, the formula is valid for all n.
Section 9.2 Series and Convergence
1. S1 = 1
3. S1 = 3
S2 = 1 +
1
4
S3 = 1 +
1
4
S4 = 1 +
1
4
S5 = 1 +
1
4
= 1.2500
+
1
9
≈ 1.3611
+
1
9
+
1
16
+
1
9
+
1
16
≈ 1.4236
+
1
25
≈ 1.4636
S2 = 3 −
9
2
= −1.5
S3 = 3 −
9
2
+
27
4
= 5.25
S4 = 3 −
9
2
+
27
4
−
81
8
= −4.875
S5 = 3 −
9
2
+
27
4
−
81
8
+
S2 = 1 +
1
3
≈ 1.3333
S3 = 1 +
1
3
+
1
5
≈ 1.5333
S4 = 1 +
1
3
+
1
5
+
1
7
≈ 1.6762
S5 = 1 +
1
3
+
1
5
+
1
7
+
243
16
= 10.3125
4. S1 = 1
2. S1 =
1
6
≈ 0.1667
S2 =
1
6
+
1
6
≈ 0.3333
S3 =
1
6
+
1
6
+
S4 =
1
6
+
1
6
S5 =
1
6
+
1
6
3
20
≈ 0.4833
+
3
20
+
2
15
≈ 0.6167
+
3
20
+
2
15
+
5
42
≈ 0.7357
1
9
≈ 1.7873
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
5. S1 = 3
∞
12.
S2 = 3 +
3
2
= 4.5
S3 = 3 +
3
2
+
3
4
= 5.250
S4 = 3 +
3
2
+
3
4
+
3
8
= 5.625
S5 = 3 +
3
2
+
3
4
+
3
8
+
3
16
r = 1.03 > 1
Diverges by Theorem 9.6
= 5.8125
= 0.5
lim
1
2
+
1
6
≈ 0.6667
S4 = 1 −
1
2
+
1
6
−
1
24
≈ 0.6250
S5 = 1 −
1
2
+
1
6
−
1
24
+
1
120
≈ 0.6333
14.
diverges
lim
∞
( 54 ) + 3( 54 )
∑ an
= 3
n =1
(
9.
n2
n +1
2
lim
n→∞
n2
=1 ≠ 0
n +1
2
Diverges by Theorem 9.9
3
2
( 54 )
+3
Geometric series, r =
4
5
3
+
16.
∞
n
∑
n2 + 1
n
n =1
converges
lim
)
n2 + 1
n→∞
<1
∑ ( 76 )
r =
∞
17.
7
6
>1
∞
∑ 5(1011 )
n→∞
n
2n + 1
1 + 2− n
1
= lim
=
≠ 0
n +1
n→∞
2
2
2
Diverges by Theorem 9.9
18.
∞
∑
n =1
Geometric series
r =
2n + 1
2n + 1
lim
n=0
11
10
∑
n =1
Diverges by Theorem 9.6
lim
n!
2n
n→∞
>1
n!
= ∞
2n
Diverges by Theorem 9.9
Diverges by Theorem 9.6
∞
11.
∑
=1≠ 0
1 + (1 n 2 )
n→∞
n
Geometric series
∞
1
= lim
Diverges by Theorem 9.9
∞
n=0
10.
∑
n =1
n
∞
n
1
=
≠ 0
2n + 3
2
Diverges by Theorem 9.9
n
2
n
2n + 3
n→∞
15.
( 54 )
{a } = {3( 54 ), 3( 54 ) , 3( 54 ) , …} converges to 0
an = 3
8.
∞
∑
n =1
n +1
n
⎧2 3 4 ⎫
= ⎨ , , , …⎬ converges to 1
⎩1 2 3 ⎭
2
3
4
∑ an = 1 + 2 + 3 +
n =1
n
=1≠ 0
n +1
Diverges by Theorem 9.9
an =
∞
n
n +1
n→∞
S3 = 1 −
{an}
∞
∑
n =1
1
2
n
Geometric series
13.
S2 = 1 −
273
n=0
6. S1 = 1
7.
∑ 2(−1.03)
Series and Convergence
1000(1.055)
n
n=0
Geometric series
19.
∑ 94 ( 14 )
n
=
n=0
S0 =
9
,S
4 1
9⎡
1
4⎣
=
9
4
+
⋅
5
4
1
4
+
=
1
16
45
,
16
+
S2 =
⎤
⎦
9
4
⋅
21
16
≈ 2.95, …
r = 1.055 > 1
Matches graph (c). Analytically, the series is geometric:
Diverges by Theorem 9.6
∑ ( 94 )( 14 )
∞
n=0
n
=
94
1 −1 4
=
94
34
= 3
© 2010 Brooks/Cole, Cengage Learning
274
20.
Chapter 9
∞
∑ ( 23 )
n
Infinite Series
=1+
n=0
S0 = 1, S1 =
2 4
+ +
3 9
24.
∑ ( 23 )
n
=
n=0
21.
∞
∑
5
, S 2 ≈ 2.11, ….
3
n=0
1
4
+
1
16
∑
n=0
22.
−
∞
⎤
⎦
25.
5
6
<1
Converges by Theorem 9.6
∞
26.
∑ 2(− 12 )
n
n=0
17 ⎛ 8 ⎞
17 ⎡
8 64
⎜− ⎟ =
⎢1 − 9 + 81 −
3
9
3
⎝
⎠
⎣
n=0
⎤
⎥
⎦
Geometric series with r = − 12 < 1
Converges by Theorem 9.6
17
, S1 ≈ 0.63, S3 ≈ 5.1, …
3
∞
27.
Matches (d). Analytically, the series is geometric:
17 ⎛ 8 ⎞
17 3
17 3
∑ 3 ⎜⎝ − 9 ⎟⎠ = 1 − (−8 9) = 17 9 = 3
n=0
n
17 ⎛ 1 ⎞
17 ⎡
1 1
1− + −
⎜− ⎟ =
3 ⎝ 2⎠
3 ⎣⎢
2 4
∑ (0.9)
n
n=0
n
∞
n=0
n
Geometric series with r =
n
∑
∞
∑ ( 56 )
n
∞
∑
n
n=0
15
45
, S1 =
, S 2 ≈ 3.05, …
4
16
15 ⎛ 1 ⎞
15 4
15 4
=
= 3
⎜− ⎟ =
4 ⎝ 4⎠
1 − ( −1 4)
54
S0 =
23.
7
,…
5
1
5
⎛ 2⎞
∑ ⎜ ⎟ = 1 − (2 5) = 3
n=0 ⎝ 5 ⎠
Matches graph (a). Analytically, the series is geometric:
∞
2
4
+
+
5
25
Matches (e). The series is geometric:
∞
n
S0 =
=1+
S0 = 1, S1 =
1
1
=
= 3
1− 23
13
15 ⎛ 1 ⎞
15
⎡1 −
⎜− ⎟ =
4 ⎝ 4⎠
4⎣
n
n=0
Matches graph (b). Analytically, the series is geometric:
∞
∞
∑ ( 52 )
Geometric series with r = 0.9 < 1
Converges by Theorem 9.6
∞
⎤
⎥
⎦
28.
∑ (−0.6)
n
n=0
17
17
, S1 =
,…
S0 =
3
6
Geometric series with r = −0.6 < 1
Converges by Theorem 9.6
Matches (f ). The series is geometric:
n
∞
17 ⎛ 1 ⎞
17
1
17 2
34
=
=
⎜− ⎟ =
3 1 − ( −1 2)
3 3
9
n=0 3 ⎝ 2 ⎠
∑
∞
29.
1
∑ n(n + 1)
n =1
∞
1
∞
1
∑ n(n + 1)
n =1
30.
∑ n(n + 2)
n =1
∞
1
∑
n = 1 n( n + 2)
31. (a)
∞
=
∞
⎛1
∑ ⎜⎝ n
1 ⎞ ⎛
1 ⎞ ⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞
⎟ = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
2 ⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠ ⎝ 4 5⎠
n + 1⎠ ⎝
−
n =1
, Sn = 1 −
1
n +1
1 ⎞
⎛
= lim S n = lim ⎜1 −
⎟ =1
n→∞
n→∞ ⎝
n + 1⎠
=
∞
⎛ 1
∑ ⎜⎜ 2n
n =1
⎝
−
⎞ ⎛1
1
1⎞ ⎛ 1 1⎞ ⎛1
1 ⎞ ⎛1
1⎞ ⎛1
1⎞
−
⎟ = ⎜ − ⎟ +⎜ − ⎟ + ⎜ −
⎟ + ⎜ −
⎟ + ⎜
⎟+
2( n + 2) ⎟⎠ ⎝ 2 6 ⎠ ⎝ 4 8 ⎠ ⎝ 6 10 ⎠ ⎝ 8 12 ⎠ ⎝ 10 14 ⎠
⎡1
⎤
1
1
1
1
1
3
= lim S n = lim ⎢ + −
−
+
=
⎥ =
n→∞
n→∞ ⎢2
n
n
4
2
1
2
2
2
4
4
+
+
(
)
(
)
⎥⎦
⎣
6
∑ n(n + 3)
n =1
∞
1 ⎞
⎡⎛
1⎞ ⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞
⎛1
= 2∑ ⎜ −
⎟ = 2 ⎢⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
n
n
3
4⎠ ⎝ 2 5⎠ ⎝3 6⎠ ⎝ 4 7⎠
+
⎠
⎣⎝
n =1 ⎝
⎤
⎥
⎦
⎛
1 1 ⎛ 1
1
1 ⎞⎤ ⎞
1 1 ⎞ 11
⎡
⎛
+
+
≈ 3.667
⎜ S n = 2 ⎢1 + + − ⎜
⎟⎥ ⎟ = 2⎜1 + + ⎟ =
2 3 ⎝ n + 1 n + 2 n + 3 ⎠⎦ ⎠
2 3⎠
3
⎝
⎣
⎝
(b)
n
5
10
20
50
100
Sn
2.7976
3.1643
3.3936
3.5513
3.6078
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
(c)
Series and Convergence
275
5
0
11
0
(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.
32. (a)
∞
4
∑ n(n + 4)
n =1
=
∞
⎛1
∑ ⎜⎝ n
1 ⎞
⎟
n + 4⎠
−
n =1
⎛
1 ⎞ ⎛1
1 ⎞ ⎛1
1 ⎞ ⎛1
1⎞ ⎛1
1⎞ ⎛1
1 ⎞
= ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ + ⎜ −
⎟ +
5
2
6
3
7
4
8
5
9
6
10
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
1 1 1
25
=1+ + +
=
≈ 2.0833
2 3 4
12
(b)
(c)
n
5
10
20
50
100
Sn
1.5377
1.7607
1.9051
2.0071
2.0443
4
0
11
0
(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.
33. (a)
∞
∑ 2(0.9)
n −1
=
n =1
(b)
(c)
∞
∑ 2(0.9)
n
=
n=0
2
= 20
1 − 0.9
n
5
10
20
50
100
Sn
8.1902
13.0264
17.5685
19.8969
19.9995
22
0
11
0
(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.
34. (a)
∞
∑ 3(0.85)
n −1
=
n =1
(b)
(c)
3
= 20
1 − 0.85
(Geometric series)
n
5
10
20
50
100
Sn
11.1259
16.0625
19.2248
19.9941
19.999998
24
0
11
0
(d) The terms of the series decrease in magnitude slowly. So, the sequence of partial sums approaches the sum slowly.
© 2010 Brooks/Cole, Cengage Learning
276
Chapter 9
35. (a)
Infinite Series
∞
∑ 10(0.25)
n −1
=
n =1
(b)
(c)
10
40
=
≈ 13.3333
1 − 0.25
3
n
5
10
20
50
100
Sn
13.3203
13.3333
13.3333
13.3333
13.3333
15
0
11
0
(d) The terms of the series decrease in magnitude rapidly. So, the sequence of partial sums approaches the sum rapidly.
36. (a)
∞
⎛ 1⎞
∑ 5⎜⎝ − 3 ⎟⎠
n −1
=
n =1
(b)
(c)
5
15
=
= 3.75
1 − ( −1 3)
4
n
5
10
20
50
100
Sn
3.7654
3.7499
3.7500
3.7500
3.7500
7
0
11
0
(d) The terms of the series decrease in magnitude rapidly. So, the sequence of partial sums approaches the sum rapidly.
∞
37.
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠
n
1
= 2
1 − (1 2)
=
n=0
∞
38.
⎛ 4⎞
∑ 6⎜⎝ 5 ⎟⎠
n
=
n=0
6
= 30
1 − ( 4 5)
(Geometric)
∞
39.
⎛ 1⎞
∑ ⎜⎝ − 3 ⎟⎠
n
=
n=0
∞
40.
⎛ 6⎞
∑ 3⎜⎝ − 7 ⎟⎠
n=0
∞
41.
1
3
=
1 − (1 3)
4
n
=
3
21
=
1 − ( −6 7)
13
∑ n2
∞
1
12 ⎞
1 ∞ ⎛ 1
1 ⎞
⎛ 12
=∑ ⎜
−
−
⎟ = ∑ ⎜
⎟
− 1 n = 2 ⎝ n − 1 n + 1⎠
2 n = 2 ⎝ n − 1 n + 1⎠
Sn =
1 ⎡⎛
1⎞ ⎛ 1 1 ⎞ ⎛1 1⎞
⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
2 ⎢⎣⎝
3⎠ ⎝ 2 4 ⎠ ⎝ 3 5⎠
n=2
∞
∑ n2
n=2
1⎞ ⎛ 1
1 ⎞⎤
1⎛
1 1
1 ⎞
⎛ 1
+⎜
− ⎟+⎜
−
⎟⎥ = ⎜1 + − −
⎟
2
1
1
2
2
n
n
n
n
n
n
−
−
+
+ 1⎠
⎝
⎠ ⎝
⎠⎦
⎝
1
1⎛
1 1
1 ⎞
3
= lim S n = lim ⎜1 + − −
⎟ =
n →∞
n →∞ 2 ⎝
−1
2 n n + 1⎠
4
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
42.
∞
n =1
⎡⎛
1⎞ ⎛ 1 1⎞ ⎛1 1⎞
S n = 2 ⎢⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
3⎠ ⎝ 2 4⎠ ⎝ 3 5⎠
⎣⎝
n =1
∞
8
∑ (n + 1)(n + 2)
n =1
∞
1 ⎞
⎛ 1
= 8∑ ⎜
−
⎟
1
2⎠
n
n
+
+
⎝
n =1
⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
S n = 8⎢⎜ − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
⎣⎝ 2 3 ⎠ ⎝ 3 4 ⎠ ⎝ 4 5 ⎠
∞
n =1
∞
1
∑ (2n + 1)(2n + 3)
n =1
Sn =
1
∑ (2n + 1)(2n + 3)
n =1
∞
45.
=
1 ∞ ⎛ 1
1
−
∑⎜
2 n = 1 ⎝ 2 n + 1 2n +
1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞
⎜ − ⎟+⎜ − ⎟+⎜ − ⎟+
2 ⎣⎢⎝ 3 5 ⎠ ⎝ 5 7 ⎠ ⎝ 7 9 ⎠
∞
⎛1⎞
∑ ⎜⎝ 10 ⎟⎠
n
1 ⎞⎤
1 ⎞
⎛ 1
⎛1
+⎜
−
⎟⎥ = 8⎜ −
⎟
⎝ n + 1 n + 2 ⎠⎦
⎝ 2 n + 2⎠
1 ⎞
⎛1
= lim Sn = lim 8⎜ −
⎟ = 4
n→∞
n→∞ ⎝ 2
n + 2⎠
8
∑ (n + 1)(n + 2)
44.
1 ⎞ ⎛1
1 ⎞⎤
1
1
1 ⎞
⎛ 1
⎛
+⎜
−
−
⎟+⎜ −
⎟⎥ = 2⎜1 + −
⎟
2 n + 1 n + 2⎠
⎝ n − 1 n + 1 ⎠ ⎝ n n + 2 ⎠⎦
⎝
1
1
1 ⎞
⎛
= lim Sn = lim 2⎜1 + −
−
⎟ = 3
n →∞
n →∞ ⎝
2 n + 1 n + 2⎠
4
∑ n ( n + 2)
43.
= lim S n = lim
n →∞
n →∞
⎞
⎟
3⎠
1
⎛ 1
+⎜
−
+
+
2
1
2
n
n
⎝
1⎛ 1
1
⎞⎤
⎟ = ⎜ −
3 ⎠⎦⎥
2 ⎝ 3 2n +
⎞
⎟
3⎠
1⎛ 1
1 ⎞
1
⎜ −
⎟ =
2 ⎝ 3 2n + 3 ⎠
6
∞
1
10
=
1 − (1 10)
9
=
n=0
50.
∑ ⎡⎣(0.7)
n
n =1
n
+ (0.9) ⎤ =
⎦
n
∞
8
⎛ 3⎞
46. ∑ 8⎜ ⎟ =
= 32
4
1
−
(3 4)
⎝
⎠
n=0
∞
47.
⎛ 1⎞
∑ 3⎜⎝ − 3 ⎟⎠
277
∞
1 ⎞
⎛1
= 2∑ ⎜ −
⎟
n + 2⎠
n =1 ⎝ n
4
∑ n ( n + 2)
∞
Series and Convergence
n
3
9
=
1 − ( −1 3)
4
=
n=0
∞
⎛7⎞
∑ ⎜⎝ 10 ⎟⎠
n
+
n=0
∞
⎛9⎞
∑ ⎜⎝ 10 ⎟⎠
n
− 2
n=0
=
1
1
+
− 2
1 − (7 10) 1 − (9 10)
=
10
34
+ 10 − 2 =
3
3
51. Note that sin (1) ≈ 0.8415 < 1. The series
∞
∑ ⎡⎣sin(1)⎤⎦
n
n =1
∞
48.
⎛ 1⎞
∑ 4⎜⎝ − 2 ⎟⎠
n
4
8
=
1 − ( −1 2)
3
=
n=0
∞
49.
is geometric with r = sin (1) < 1. So,
n
n
∞
⎛1⎞
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ − ∑ ⎜⎝ 3 ⎟⎠
n=0
n=0
1
1
3
1
=
−
= 2− =
1 − (1 2) 1 − (1 3)
2
2
1⎞
⎛1
∑ ⎜⎝ 2n − 3n ⎟⎠ =
n=0
52. S n =
n
∑ 9k 2
k =1
∞
∞
∑ ⎣⎡sin(1)⎦⎤
1
=
+ 3k − 2
n
∑ (3k
k =1
n =1
n
∞
= sin (1) ∑ ⎡⎣sin (1)⎤⎦ =
n=0
n
sin (1)
≈ 5.3080.
1 − sin (1)
1
− 1)(3k + 2)
n
1 ⎤
1 n ⎡ 1
1 ⎤
⎡ 1
= ∑⎢
−
= ∑⎢
−
⎥
9k + 6 ⎦
3 k =1 ⎣ 3k − 1 3k + 2 ⎥⎦
k =1 ⎣ 9 k − 3
=
1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1
1⎞
⎜ − ⎟+⎜ − ⎟+⎜ − ⎟+
3 ⎢⎣⎝ 2 5 ⎠ ⎝ 5 8 ⎠ ⎝ 8 11 ⎠
lim Sn = lim
n→∞
n→∞
1 ⎞⎤
1⎛ 1
1 ⎞
⎛ 1
+⎜
−
⎟⎥ = ⎜ −
⎟
3 ⎝ 2 3n + 2 ⎠
⎝ 3n − 1 3n + 2 ⎠⎦
1⎛ 1
1 ⎞
1
⎜ −
⎟ =
3 ⎝ 2 3n + 2 ⎠
6
© 2010 Brooks/Cole, Cengage Learning
278
Chapter 9
53. (a) 0.4 =
Infinite Series
∞
4⎛1⎞
∑ 10 ⎜⎝ 10 ⎟⎠
n
58. (a) 0.215 =
n=0
(b) Geometric series with a =
S =
4
1
and r =
10
10
n
∞
⎛ 1 ⎞
∑ ⎜⎝ 100 ⎟⎠
n
=
n =1
(b) 0.01 =
3⎛ 1 ⎞
57. (a) 0.075 = ∑ ⎜
⎟
n = 0 40 ⎝ 100 ⎠
S =
∞
63.
⎛1
∑ ⎜⎝ n
−
n =1
81
1
and r =
100
100
∞
⎛1
−
n =1
∞
64.
⎛
lim
n→∞
n + 10
1
=
≠ 0
10n + 1 10
Diverges by Theorem 9.9
∞
62.
4n + 1
∑ 3n − 1
n =1
lim
n
n→∞
4n + 1
4
=
≠ 0
3n − 1
3
Diverges by Theorem 9.9
3
1
and r =
40
100
1 ⎞ ⎛1
1 ⎞
1
1
1
⎛ 1
+⎜
−
−
⎟+⎜ −
⎟ = 1+ −
2 n+1 n + 2
⎝ n − 1 n + 1⎠ ⎝ n n + 2 ⎠
1 ⎞
⎟
n + 2⎠
⎛ 1 1⎞ ⎛1 1⎞
Sn = ⎜ − ⎟ + ⎜ − ⎟ +
⎝ 2 3⎠ ⎝ 3 4⎠
∑ ⎜⎝ n + 1 −
n =1
n =1
1 ⎞
1
1
1 ⎞
3
⎛
S n = lim ⎜1 + −
−
⎟ = nlim
⎟ = , converges
→
∞
→
∞
n
2
2
n + 2⎠
n + 1 n + 2⎠
⎝
1
1
n + 10
∑ 10n + 1
1 ⎞
⎟
n + 2⎠
n =1
⎛
∞
61.
a
3 40
5
=
=
1− r
99 100
66
∑ ⎜⎝ n + 1 −
∞
Diverges by Theorem 9.6
n
1⎞ ⎛ 1 1 ⎞ ⎛1 1⎞
⎛
S n = ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
3⎠ ⎝ 2 4⎠ ⎝ 3 5⎠
⎝
∑ ⎜⎝ n
3n
∑ 1000
n =1
1 ∞ ⎛ 1 ⎞
∑⎜ ⎟
100 n = 0 ⎝ 100 ⎠
(b) Geometric series with a =
n
Geometric series with r = 3 > 1.
1
1
1 100
1
⋅
=
⋅
=
100 1 − (1 100)
100 99
99
∞
∑ (1.075)
∞
60.
a
81 100
81
9
=
=
=
S =
1− r
1 − (1 100)
99
11
56. (a) 0.01 =
3
1
and r =
200
100
Diverges by Theorem 9.6
n
(b) Geometric series with a =
n=0
Geometric series with r = 1.075
9
1
=1
10 1 − (1 10)
81 ⎛ 1 ⎞
55. (a) 0.81 = ∑
⎜
⎟
n = 0 100 ⎝ 100 ⎠
n
n=0
9 ∞ ⎛1⎞
⎛1⎞
= ∑ 9⎜ ⎟ =
∑⎜ ⎟
10 n = 0 ⎝ 10 ⎠
n =1 ⎝ 10 ⎠
∞
3 ⎛ 1 ⎞
1
a
1
3 200
71
+
= +
=
5 1− r
5 99 100
330
∞
59.
n
∞
(b) 0.9 =
S =
9
9
9
+
+
+
10 100 1000
∞
∑ 200 ⎜⎝ 100 ⎟⎠
(b) Geometric series with a =
a
4 10
4
=
=
1− r
1 − (1 10)
9
54. (a) 0.9 =
1
+
5
1 ⎞
1
1
⎛ 1
+⎜
−
−
⎟ =
2 n+ 2
⎝ n + 1 n + 2⎠
1 ⎞
1 ⎞
1
⎛1
Sn = lim ⎜ −
⎟ = nlim
⎟ = , converges
→∞
n→∞ ⎝ 2
2
n + 2⎠
n + 2⎠
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
65.
∞
1
∑ n(n + 3)
=
n =1
Sn =
=
∞
1⎛
1 1
1
1
1 ⎞
1 ⎛ 11 ⎞
11
= lim S n = lim ⎜1 + + −
−
−
⎟ = ⎜ ⎟ = , converges
n →∞
n →∞ 3 ⎝
2 3 n + 1 n + 2 n + 3⎠
3⎝ 6 ⎠
8
n =1
1
∑ 2n(n + 1)
=
n =1
Sn =
∞
1
= lim S n = lim
n =1
∞
1 ∞ ⎛1
1 ⎞
∑ ⎜ − n + 1 ⎟⎠
2 n =1 ⎝ n
1 ⎡⎛
1 ⎞ ⎛ 1 1⎞ ⎛1 1⎞
⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
2 ⎢⎣⎝
2 ⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠
∑ 2n(n + 1)
67.
1 ⎞ ⎛ 1
1 ⎞ ⎛1
1 ⎞⎤
⎛ 1
+ ⎜
−
−
⎟ −⎜
⎟ −⎜ −
⎟
n + 1⎠ ⎝ n − 1 n + 2 ⎠ ⎝ n
n + 3 ⎠⎥⎦
⎝n − 2
1⎛
1
1
1
1
1 ⎞
−
−
⎜1 + + −
⎟
3⎝
2
3 n +1 n + 2
n + 3⎠
1
∞
n→∞
n→∞
1 ⎞⎤
1⎛
1 ⎞
⎛1
+⎜ −
⎟⎥ = ⎜1 −
⎟
n + 1⎠
2⎝
⎝ n n + 1 ⎠⎦
1⎛
1 ⎞
1
⎜1 −
⎟ = , converges
2⎝
2
n + 1⎠
3n − 1
∑ 2n + 1
72. S n =
n =1
68.
n
3
= lim
(ln 2)2 3n
n→∞
n=0
6n
= lim
(ln n)3 3n
n→∞
6
∞
∑
n=0
n =1
⎛1⎞
= 4∑ ⎜ ⎟
n=0 ⎝ 2 ⎠
n
n→∞
So,
∞
⎡
∑ ⎢⎣1 +
n =1
∞
74.
n =1
r =
n
∞
n
⎛1⎞
∑ ⎜⎝ e ⎟⎠ converges because it is geometric
n =1
1
< 1.
e
75. lim arctan n =
1
Geometric series with r =
5
approach 0 as n → ∞. So, the series
∑ e− n =
n
k⎤
diverges.
n ⎥⎦
with
∞
3
⎛1⎞
=
3
∑
⎜ ⎟ , convergent
n
S
n=0 ⎝ 5⎠
71. Because n > ln ( n), the terms an =
k
For k = 0, lim (1 + 0) = 1 ≠ 0.
n
Converges by Theorem 9.6
70.
⎛1⎞
∑ ln⎜⎝ n ⎟⎠ diverges.
= e k ≠ 0.
= ∞
1
Geometric series with r =
2
∞
− ln ( n)
∞
n
nk
⎡⎛
k⎞
k⎞ ⎤
⎛
lim ⎜1 + ⎟ = lim ⎢⎜1 + ⎟ ⎥
n→∞ ⎝
n→∞ ⎝
n⎠
n ⎠ ⎦⎥
⎣⎢
(by L'Hôpital's Rule); diverges by Theorem 9.9
4
k =1
73. For k ≠ 0,
(ln 2)3n
3n
lim 3 = lim
n→∞ n
n→∞
3n 2
∑ 2n
n
∑ − ln(k )
n→∞
n =1
69.
=
Because lim Sn diverges,
∑ n3
∞
⎛1⎞
= 0 − ln 2 − ln 3 −
Diverges by Theorem 9.9
∞
n
∑ ln⎜⎝ k ⎟⎠
k =1
3n − 1
3
=
≠ 0
2n + 1
2
lim
n→∞
279
1 ∞ ⎛1
1 ⎞
∑ ⎜ − n + 3 ⎟⎠
3 n =1 ⎝ n
1 ⎡⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1 1 ⎞ ⎛ 1
1⎞
⎜ − ⎟ +⎜ − ⎟ +⎜ − ⎟ +⎜ − ⎟ +
3 ⎢⎣⎝ 1 4 ⎠ ⎝ 2 5 ⎠ ⎝ 3 6 ⎠ ⎝ 4 7 ⎠
∑ n(n + 3)
66.
Series and Convergence
n→∞
n
do not
ln ( n)
∞
So,
π
2
≠ 0
∞
∑ arctan n diverges.
n =1
n
∑ ln(n) diverges.
n=2
© 2010 Brooks/Cole, Cengage Learning
280
Chapter 9
Infinite Series
∞
⎛ k + 1⎞
⎟
k ⎠
n
∑ ln⎜⎝
76. S n =
k =1
84.
∑ (3 x )
n =1
⎛ n + 1⎞
+ ln⎜
⎟
⎝ n ⎠
⎛ 2⎞
⎛ 3⎞
= ln⎜ ⎟ + ln⎜ ⎟ +
⎝1⎠
⎝ 2⎠
∞
Diverges
85.
∑ ( x − 1)
∞
= ( x − 1) ∑ ( x − 1)
n
n =1
77. See definitions on page 608.
n→∞
5.
∞
∑ an
= a1 + a2 +
f ( x ) = ( x − 1) ∑ ( x − 1)
1
x −1
= ( x − 1)
=
,
1 − ( x − 1)
2− x
partial sums is 5.
79. The series given by
∑ ar n
∞
= a + ar + ar 2 +
+ ar n +
,a ≠ 0
86.
n=0
r ≥ 1.
n→∞
4
⎛ x − 3⎞
⎟ =
4
−
⎡
x
1
(
⎠
n=0
⎣ − 3) 4⎤⎦
4
16
=
=
, −1 < x < 7
(4 − x + 3) 4 7 − x
∞
∑ an diverges.
n =1
∞
∑ an
∞
= a1 + a2 + a3 +
87.
n =1
∑ (−1)
n
xn =
n=0
∞
∑ ak
∞
∑ ( − x)
n=0
− x < 1 ⇒ x < 1 ⇒ −1 < x < 1
These are the same. The third series is different,
unless a1 = a2 =
= a is constant.
= ak + ak +
∞
∑ (− x)
f ( x) =
∞
∑ (−1)
n
x 2n =
n=0
82. (a) Yes, the new series will still diverge.
∞
n
x ∞ ⎛ x⎞
⎛ x⎞
∑ ⎜⎝ 2 ⎟⎠ = 2 ∑ ⎜⎝ 2 ⎟⎠
n =1
n=0
x ∞ ⎛ x⎞
f ( x) = ∑ ⎜ ⎟
2 n=0 ⎝ 2 ⎠
=
∞
f ( x) =
∑ (− x 2 )
−1 < x < 1
n
n=0
∞
∑ (− x 2 )
n
n=0
x
< 1 or x < 2
2
∞
89.
⎛1⎞
∑ ⎜⎝ x ⎟⎠
=
1
1
=
, −1 < x < 1
2
+
1 − (− x ) 1 x 2
n
n=0
n
1
x
x 2
x
=
=
,
2 1 − ( x 2)
22 − x
2− x
1
,
1+ x
− x 2 < 1 ⇒ −1 < x < 1
n
Geometric series: converges for
=
Geometric series: converges for
(b) Yes, the new series will converge.
xn
∑ 2n =
n =1
n
n=0
88.
n =1
∞
n
Geometric series: converges for
= a1 + a2 + a3 +
k =1
∑ ak
n
∞
∑ 4⎜⎝
f ( x) =
80. If lim an ≠ 0, then
∞
n
x −3
< 1 ⇒ x − 3 < 4 ⇒ −1 < x < 7
4
series converges to a (1 − r ). The series diverges if
(b)
⎛ x − 3⎞
∑ 4⎜ 4 ⎟⎠
n=0 ⎝
0 < x < 2
Geometric series: converges for
is a geometric series with ratio r. When 0 < r < 1, the
81. (a)
n
n=0
= 5 means that the limit of the
n =1
∞
n
n=0
Geometric series: converges for
x −1 <1⇒ 0 < x < 2
78. lim an = 5 means that the limit of the sequence {an} is
∞
1
3
∞
1
3x
1
n
=
f ( x ) = (3 x ) ∑ (3 x ) = (3 x )
, x <
−
−
1
3
x
1
3
x
3
n=0
= ln(n + 1) − ln(1) = ln(n + 1)
83.
n
n=0
Geometric series: converges for 3 x < 1 ⇒ x <
+ (ln(n + 1) − ln n)
= (ln 2 − ln1) + (ln 3 − ln 2) +
(c)
∞
= (3 x ) ∑ (3 x )
n
Geometric series: converges if
x < 2
1
<1
x
⇒ x > 1 ⇒ x < −1 or x > 1
f ( x) =
∞
⎛1⎞
∑ ⎜⎝ x ⎟⎠
n=0
n
=
1
x
, x > 1 or x < −1
=
1 − (1 x)
x −1
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
∞
90.
⎛
∑ ⎜ x2
n =1 ⎝
n
x2 ⎞
x2 ∞ ⎛ x2 ⎞
⎟ = 2
⎟
∑⎜
+ 4⎠
x + 4 n = 0 ⎝ x2 + 4 ⎠
94. (a) False. The fact that
x2
⋅
x2 + 4
∞
91.
∑ (1 + c)
−n
1
1−
x2
x + 4
∑ (1 + c)
n=2
x2 x2 + 4
x2
=
4
x2 + 4 4
95. (a) x is the common ratio.
−n − 2
n=0
=
∞
⎛ 1 ⎞
⎟
2 ∑⎜
(1 + c) n = 0 ⎝ 1 + c ⎠
1
1
<1⇒1< 1+ c
1+ c
For convergence,
⇒ c > 0 or c < −2
∞
1
2 =
(1 + c)
2
⎛ 1 ⎞
⎟
+ c⎠
∑⎜
n
n = 0 ⎝1
1+ c
2(1 + c) =
=
c
⎛ 1 ⎞
1−⎜
⎟
⎝1 + c ⎠
1
2(1 + c) =
c
2c 2 + 2c − 1 = 0
c =
−2 +
4+8
92.
∞
−1 +
2
∑ ecn
n=0
−1 ± 3
=
2
4
Because −2 <
c =
=
n
(b) 1 + x + x 2 +
−1 −
2
3
=
∞
∑ xn
(c) y1 =
n
y3 = S5 = 1 + x + x 2 + x 3 + x 4
Answers will vary.
3
f
S5
S3
− 1.5
1.5
0
⎛ x⎞
96. (a) ⎜ − ⎟ is the common ratio.
⎝ 2⎠
(b) 1 −
< 0, the answer is
=5
n=0
1
= 5
1 − ec
1
1 − ec =
5
4
ec =
5
x <1
y2 = S 3 = 1 + x + x 2
x
x2
x3
+
−
+
2
4
8
(c) y1 =
∞
n
⎛ x⎞
∑ ⎜− ⎟
n=0 ⎝ 2 ⎠
1
=
1 − ( − x 2)
=
=
∑ (e c )
1
,
1− x
1
1− x
3
∞
=
n=0
1
2
4
2
∞
=
=
1
→ 0 is irrelevant to the
n
1
1
convergence of ∑ 4 . In fact, ∑ 4 diverges.
n
n
(b) False. The fact that
⇒ x 2 < x 2 + 4 ⇒ converges for all x
f ( x) =
281
1
→ 0 is irrelevant to the
n4
∞
∞
1
1
convergence of ∑ 4 . Furthermore, ∑ 4 ≠ 0.
n
n
n =1
n =1
n
x2
<1
x2 + 4
Geometric series: converges for
Series and Convergence
2
,
2+ x
x < 2
2
2+ x
x
x2
+
2
4
x
x2
x3
x4
y3 = S5 = 1 − +
−
+
2
4
8
16
y2 = S 3 = 1 −
Answers will vary.
⎛4⎞
c = ln ⎜ ⎟
⎝5⎠
5
f
S5
S3
−5
5
93. Neither statement is true. The formula
1
= 1 + x + x2 +
1− x
−5
holds for −1 < x < 1.
© 2010 Brooks/Cole, Cengage Learning
282
Chapter 9
Infinite Series
⎡1 − 0.5 x ⎤
97. f ( x) = 3⎢
⎥
⎣ 1 − 0.5 ⎦
100.
Horizontal asymptote: y = 6
∞
⎛1⎞
∑ 3⎜⎝ 2 ⎟⎠
n=0
This inequality is true when n = 14.
n
(0.01)
< 0.0001
This inequality is true when n = 5. This series
converges at a faster rate.
The horizontal asymptote is the sum of the series.
f ( n) is the nth partial sum.
n −1
101.
y=6
7
n
10,000 < 10n
3
= 6
1 − (1 2)
S =
1
< 0.0001
2n
10,000 < 2n
8000 ⎡⎣1 − 0.95n ⎤⎦
1 − 0.95
∑ 8000(0.95) =
i
i =0
= 160,000 ⎡⎣1 − 0.95n ⎤⎦ , n > 0
102. V (t ) = 475,000(1 − 0.3) = 475,000(0.7)
n
−2
10
V (5) = 475,000(0.7) = $79,833.25
5
−1
∞
⎡1 − 0.8 x ⎤
98. f ( x) = 2 ⎢
⎥
⎣ 1 − 0.8 ⎦
103.
()
n
4
5
S =
1
= 10
1 − ( 4 5)
n=0
= 800 million dollars
∑ 200(0.60)
= 500 million dollars
105. D1 = 16
D2 = 0.81(16) + 0.81(16) = 32(0.81)
y = 10
up
down
D3 = 16(0.81) + 16(0.81) = 32(0.81)
2
2
2
∞
∑ 32(0.81)
n
= −16 +
n=0
−2
16
106. The ball in Exercise 105 takes the following times for
each fall.
1
< 0.0001
n( n + 1)
s1 = −16t 2 + 16
10,000 < n 2 + n
s1 = 0 if t = 1
s2 = −16t + 16(0.81)
2
0 < n 2 + n − 10,000
−1 ± 12 − 4(1)( −10,000)
2
Choosing the positive value for n you have
n ≈ 99.5012. The first term that is less than 0.0001 is
n = 100.
n
⎛1⎞
⎜ ⎟ < 0.0001
⎝8⎠
10,000 < 8
32
1 − 0.81
≈ 152.42 feet
−2
n =
2
D = 16 + 32(0.81) + 32(0.81) +
= −16 +
99.
i
i=0
The horizontal asymptote is the sum of the series.
f ( n) is the nth partial sum.
12
i
∞
104.
∑2
∑ 200(0.75)
i=0
Horizontal asymptote: y = 10
∞
n
n
This inequality is true when n = 5. This series
converges at a faster rate.
s2 = 0 if t = 0.9
s3 = −16t + 16(0.81)
2
s3 = 0 if t = (0.9)
2
sn = −16t 2 + 16(0.81)
n −1
sn = 0 if t = (0.9)
n −1
2
Beginning with s2 , the ball takes the same amount of
time to bounce up as it takes to fall. The total elapsed
time before the ball comes to rest is
∞
∞
t = 1 + 2∑ (0.9) = −1 + 2 ∑ (0.9)
n =1
= −1 +
n
n
n=0
2
= 19 seconds.
1 − 0.9
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
107. P( n) =
P ( 2) =
1⎛ 1 ⎞
⎜ ⎟
2⎝ 2 ⎠
n
108. P( n) =
2
1⎛ 1 ⎞
1
⎜ ⎟ =
2⎝ 2 ⎠
8
P ( 2) =
n
∞
283
n
2
1⎛ 2 ⎞
4
⎜ ⎟ =
3⎝ 3 ⎠
27
n
∞
1⎛ 1 ⎞
12
∑ ⎜ ⎟ = 1 − (1 2) = 1
n=0 2⎝ 2 ⎠
1⎛ 2 ⎞
⎜ ⎟
3⎝ 3 ⎠
Series and Convergence
1⎛ 2 ⎞
13
∑ ⎜ ⎟ = 1 − (2 3) = 1
n = 0 3⎝ 3 ⎠
109. (a)
n
∞
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ =
n =1
∞
n
1⎛ 1 ⎞
1
1
∑ ⎜ ⎟ = 2 (1 − (1 2)) = 1
n=0 2⎝ 2 ⎠
(b) No, the series is not geometric.
(c)
110. Person 1:
1
1
1
+ 4 + 7 +
2 2
2
⎛1⎞
∑ n⎜⎝ 2 ⎟⎠ = 2
n =1
n
1 ∞ ⎛1⎞
1
1
4
=
∑
⎜ ⎟ =
2 n=0 ⎝ 8 ⎠
2 1 − (1 8)
7
=
n
Person 2:
1
1
1
+ 5 + 8 +
22
2
2
=
1 ∞ ⎛1⎞
1
1
2
∑ ⎜ ⎟ = 4 1 − (1 8) = 7
4 n=0 ⎝ 8 ⎠
Person 3:
1
1
1
+ 6 + 9 +
23
2
2
=
1 ∞ ⎛1⎞
1
1
1
∑ ⎜ ⎟ = 8 1 − (1 8) = 7
8 n=0 ⎝ 8 ⎠
Sum:
n
∞
n
4 2 1
+ +
=1
7 7 7
111. (a) 64 + 32 + 16 + 8 + 4 + 2 = 126 in.2
(b)
∞
n
64
⎛1⎞
∑ 64⎜⎝ 2 ⎟⎠ = 1 − (1 2) = 128 in.2
n=0
16 in.
Note: This is one-half of the area of the original square
16 in.
112. (a) sin θ =
sin θ =
sin θ =
Yy1
z
x1 y1
Yy1
x1 y2
x1 y1
⇒ Yy1 = z sin θ
⇒ x1 y1 = Yy1 sin θ = z sin 2 θ
⇒ x1 y2 = x1 y1 sin θ = z sin 3 θ
Total: z sin θ + z sin 2 θ + z sin 3 θ +
(b) If z = 1 and θ =
20
113.
1
)
∑ 100,000(1.06
n =1
n
π
6
= z
sin θ
1 − sin θ
12
= 1.
1 − (1 2)
, then total =
i
=
100,000 19 ⎛ 1 ⎞
100,000 ⎡1 − 1.06−20 ⎤
∑
⎢
⎥
⎜
⎟ =
1.06 i = 0 ⎝ 1.06 ⎠
1.06 ⎣ 1 − 1.06−1 ⎦
(n
= 20, r = 1.06−1 ) ≈ $1,146,992.12
The $2,000,000 sweepstakes has a present value of $1,146,992.12. After accruing interest over the 20-year period, it attains its
full value.
2
114. Surface area = 4π (1) + 9⎛⎜ 4π
⎝
( 13 ) ⎞⎟⎠ + 9
2
2
⋅ 4π
( 19 )
2
+
= 4(π + π +
)
= ∞
© 2010 Brooks/Cole, Cengage Learning
284
Chapter 9
115. w =
Infinite Series
n −1
∑ 0.01(2)
i
=
0.01(1 − 2n )
1− 2
i=0
= 0.01( 2n − 1)
(a) When n = 29: w = $5,368,709.11
(b) When n = 30: w = $10,737,418.23
(c) When n = 31: w = $21,474,836.47
12t −1
116.
∑
n=0
n
r ⎞
⎛
P⎜1 + ⎟ =
12 ⎠
⎝
121. T = 50,000 + 50,000(1.04) +
⎡
P ⎢1 −
⎣⎢
12 t
r ⎞ ⎤
⎛
⎜1 + ⎟ ⎥
12 ⎠ ⎥⎦
⎝
r ⎞
⎛
1 − ⎜1 +
⎟
12 ⎠
⎝
r ⎞
⎛ 12 ⎞ ⎡⎛
⎛
= P⎜ − ⎟ ⎢⎜1 − ⎜1 + ⎟ ⎥
⎜
12
r
⎝
⎠ ⎢⎣⎝
⎝
⎠ ⎥⎦
∑ P(e )
r 12
=
(
P 1 − (e r 12 )
1 − er 12
n=0
117. P = 45,
12t
r = 0.03,
) = P(e
rt
− 1)
118. P = 75,
(
)
≈ $14,779.65
e0.03 12 − 1
r = 0.055,
t = 25
12( 25)
⎤
0.055 ⎞
⎛ 12 ⎞ ⎡⎛
1
+
− 1⎥ ≈ $48,152.81
(a) A = 75⎜
⎢
⎟⎜
⎟
12 ⎠
⎝ 0.055 ⎠ ⎢⎣⎝
⎥⎦
(b) A =
(
)
75 e0.055(25) − 1
e0.055 12 − 1
119. P = 100,
r = 0.04,
123. False. lim
n→∞
e r 12 − 1
12( 20)
⎤
0.03 ⎞
⎛ 12 ⎞ ⎡⎛
(a) A = 45⎜
− 1⎥ ≈ $14,773.59
⎟ ⎢⎜1 +
⎟
12 ⎠
⎝ 0.03 ⎠ ⎢⎣⎝
⎥⎦
(b) A =
120. P = 30,
(
e
0.04 12
1
= 0, but
n
125. False;
∞
∑ ar n
)
−1
r = 0.06,
≈ $5,351,516.15
1
n =1
n =1
The formula requires that the geometric series begins
with n = 0.
126. True
lim
n→∞
n
1
=
≠ 0
1000( n + 1)
1000
127. True
0.74999 … = 0.74 +
≈ $48,245.07
9
9
+
+
103 104
n
= 0.74 +
9 ∞ ⎛1⎞
∑⎜ ⎟
103 n = 0 ⎝ 10 ⎠
= 0.74 +
9
1
⋅
103 1 − (1 10)
9 10
⋅
103 9
1
= 0.74 +
= 0.75
100
= 0.74 +
≈ $91,503.32
t = 50
∞
∑ n diverges.
⎛ a ⎞
= ⎜
⎟ − a
⎝1 − r ⎠
t = 35
100 e0.04(35) − 1
n
124. True
12(35)
⎤
0.04 ⎞
⎛ 12 ⎞ ⎡⎛
− 1⎥ ≈ $91,373.09
(a) A = 100⎜
⎟ ⎢⎜1 +
⎟
12 ⎠
⎝ 0.04 ⎠ ⎢⎣⎝
⎥⎦
(b) A =
39
∑ 50,000(1.045)
n=0
t = 20
45 e0.03(20) − 1
n
⎡1 − 1.0440 ⎤
= 50,000 ⎢
⎥ ≈ $4,751,275.79
⎣ 1 − 1.04 ⎦
122. T =
12 t
⎤
r ⎞
⎛ 12 ⎞ ⎡⎛
= P⎜ ⎟ ⎢⎜1 + ⎟ − 1⎥
12 ⎠
⎝ r ⎠ ⎢⎣⎝
⎥⎦
n
39
n=0
12 t ⎤
12t −1
39
∑ 50,000(1.04)
=
+ 50,000(1.04)
128. True
12(50)
⎤
0.06 ⎞
⎛ 12 ⎞ ⎡⎛
− 1⎥ ≈ 113,615.73
(a) A = 30⎜
⎟ ⎢⎜1 +
⎟
12 ⎠
⎝ 0.06 ⎠ ⎣⎢⎝
⎥⎦
(b) A =
(
)
30 e0.06(50) − 1
e0.06 12 − 1
≈ $114,227.18
© 2010 Brooks/Cole, Cengage Learning
Section 9.2
129. By letting S0 = 0, you have
n
∑ ak
an =
−
k =1
∞
∑ an
n −1
∑ ak
∞
n =1
= S n − S n −1. So,
∑(an
− S n −1 )
∞
∑ ( Sn
− S n −1 + c − c)
n=0
∑ ⎡⎣(c − Sn −1 ) − (c − Sn )⎤⎦.
+ bn ) =
n=0
⎛1⎞
∑ ⎜⎝ c ⎟⎠can
= L. Then lim S n = L and because
n→∞
∞
0
n=0
=
∑ an
So, ∑ can diverges.
you have
lim Rn = lim ( L − S n ) = lim L − lim S n = L − L = 0.
n →∞
∞
∑ [1 − 1] =
converges. This is a contradiction since ∑ can diverged.
ak = L − S n ,
k = n +1
n →∞
=
c ≠ 0,
convergent series
∑
∞
∑ ⎡⎣1 + (−1)⎤⎦
133. Suppose, on the contrary, that ∑ can converges. Because
130. Let {S n} be the sequence of partial sums for the
Rn =
n=0
contradiction. So, ∑ ( an + bn ) diverges.
n =1
n =1
∞
∑ (−1).
∑ ( an + bn ) − ∑ an = ∑ bn would converge, which is a
∞
=
=
132. If ∑ ( an + bn ) converged, then
n =1
∑ an
∞
∑ 1 and ∑bn
=
n =1
=
∞
∑an
285
Both are divergent series.
k =1
∑ ( Sn
=
131. Let
Series and Convergence
n →∞
134. If ∑ an converges, then lim an = 0. So, lim
n→∞
n →∞
which implies that
1
∑a
n→∞
1
≠ 0,
an
diverges.
n
135. (a)
1
1
a
− an + 1
an + 2
1
−
= n+3
=
=
an +1an + 2
an + 2 a n + 3
an +1an + 2 an + 3
an +1an + 2 an + 3
an +1an + 3
(b) S n =
n
∑a
k =0
=
1
k +1ak + 3
⎤
1
1
−
ak + 2 ak + 3 ⎥⎦
k = 0 ⎣ k +1ak + 2
n
⎡
∑ ⎢a
⎡ 1
1 ⎤ ⎡ 1
1 ⎤
= ⎢
−
⎥ + ⎢a a − a a ⎥ +
a
a
a
a
1
2
2
3
2
3
3
4⎦
⎣
⎦ ⎣
∞
∑a
n=0
⎡ 1
⎤
1
1
1
1
+ ⎢
−
= 1−
⎥ = aa − a a
a
a
a
a
a
an + 3
+
+
+
+
+
+
+
1
2
2
3
1
2
2
3
2
n
n
n
n
n
n
n
⎣
⎦
⎡
⎤
1
1
= lim S n = lim ⎢1 −
⎥ =1
n→∞
n→∞
a
a
a
n +1 n + 3
n+ 2 n+3 ⎦
⎣
+ x 2 n + 2 x 2 n +1
136. S 2 n = 1 + 2 x + x 2 + 2 x3 +
= (1 + x 2 + x 4 +
+ x 2 n ) + (2 x + 2 x3 +
+ 2 x 2 n +1 ) =
n
∑ ( x2 )
k
k =0
n
+ 2 x ∑ ( x2 )
k
n=0
As n → ∞, the two geometric series converge for x < 1:
lim S2 n =
n→∞
137.
1
1 + 2x
⎛ 1 ⎞
+ 2 x⎜
=
,
2⎟
1 − x2
1 − x2
⎝1 − x ⎠
1
1
1
+ 2 + 3 +
r
r
r
=
∞
1⎛ 1 ⎞
∑ r ⎜⎝ r ⎟⎠
n=0
n
=
x <1
1r
1
=
1 − (1 r )
r −1
⎛
⎞
1
< 1⎟
⎜ since
r
⎝
⎠
This is a geometric series which converges if
1
< 1 ⇔ r > 1.
r
© 2010 Brooks/Cole, Cengage Learning
286
Chapter 9
Infinite Series
138.
1
A
B
C
=
+
+
n( n + 1)( n + 2)
n
n +1 n+ 2
1 = A( n + 1)( n + 2) + Bn ( n + 2) + Cn ( n + 1)
When n = 0,
1 = 2A ⇒ A = 1 2
When n = −1,
1 = − B ⇒ B = −1
When n = −2,
1 = 2C ⇒ C = 1 2
∞
1
∑ n(n + 1)(n + 2)
=
n =1
12
−1
12
+
+
n
n +1 n + 2
1
12 ⎞
⎛1 2 1 1 2 ⎞ ⎛1 2 1 1 2 ⎞ ⎛1 2 1 1 2 ⎞
⎛1 2
Sn = ⎜
− +
− +
− +
+ ⎜
−
+
⎟ +⎜
⎟+⎜
⎟+
⎟
2
3 ⎠ ⎝ 2
3
4 ⎠ ⎝ 3
4
4 ⎠
n + 1 n + 2⎠
⎝ 1
⎝ n
1⎞
1
12 ⎞
⎛1 2 1 1 2 ⎞ ⎛ 1 1 ⎞ ⎛ 1
⎛1 2
= ⎜
− +
+⎜
−
+
⎟ +⎜ − ⎟ +⎜ − ⎟ +
⎟
2
2 ⎠ ⎝ 3 3⎠ ⎝ 4
4⎠
n + 1 n + 2⎠
⎝ 1
⎝ n
As n → ∞, S n → 1 4
∞
1
∑ n(n + 1)(n + 2)
=
n =1
139. (a)
1
4
140. The entire rectangle has area 2 because the height is 1
1
1
and the base is 1 + + +
= 2. The squares all lie
2
4
inside the rectangle, and the sum of their areas is
y
1
2
1+
1
1
1
+ 2 + 2 +
2
2
3
4
So,
∑ n2
x
1
2
1
y
∞
1
.
< 2.
n =1
1
141. Let H represent the half-life of the drug. If a patient
receives n equal doses of P units each of this drug,
administered at equal time interval of length t, the total
amount of the drug in the patient’s system at the time the
last dose is administered is given by
Tn = P + Pe kt + Pe 2 kt +
+ Pe(n −1)kt where
1
2
x
1
2
1
k = −(ln 2) H . One time interval after the last dose is
y
administered is given by
Tn + 1 = Pe kt + Pe 2 kt + Pe3kt +
+ Penkt . Two time
intervals after the last dose is administered is given by
Tn + 2 = Pe 2 kt + Pe3kt + Pe 4 kt +
+ Pe(n + 1)kt and so
on. Because k < 0, Tn + s → 0 as s → ∞, where s is an
1
1
2
x
1
2
1
1
⎡
1
x2 ⎤
∫ 0 (1 − x) dx = ⎢⎣x − 2 ⎥⎦ = 2
0
1
(b)
Sn =
⎡ x3
1 1
1
x4 ⎤
− x 3 ) dx = ⎢ −
=
⎥ = −
3
4
3
4
12
⎣
⎦
=
1
1
(c) an =
2
2
⎡
n
x
n −1
n
∫ 0 ( x − x ) dx = ⎢ n
1
−
n +1 1
x ⎤
1
1
⎥ = −
n + 1⎦ 0 n n + 1
⎣
1
1
1
⎛
⎞ ⎛
⎞
∑ an = ⎜⎝1 − 2 ⎟⎠ + ⎜⎝ 2 − 3 ⎟⎠ + = 1
n =1
Note that the square has area = 1.
∞
142. The series is telescoping:
⎡x
1 1
1
x ⎤
= ⎢
− ⎥ =
− =
2
3
2
3
6
⎣
⎦0
2
∫ 0 ( x − x ) dx
∫ 0 (x
3 1
integer.
6k
n
∑
k =1
n
(3
k +1
⎡
∑ ⎢3k
k =1 ⎣
= 3−
− 2
k +1
)(3k
− 2k )
⎤
3k
3k + 1
− k +1
k
k +1 ⎥
− 2
− 2 ⎦
3
3n + 1
− 2n +1
n +1
3
lim Sn = 3 − 1 = 2
n→∞
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
The Integral Test and p-Series
287
143. f (1) = 0, f ( 2) = 1, f (3) = 2, f ( 4) = 4, …
⎧⎪n 2 4,
n even
In general: f ( n) = ⎨ 2
⎪⎩( n − 1) 4, n odd.
(See below for a proof of this.)
x + y and x − y are either both odd or both even. If both even, then
f ( x + y) − f ( x − y) =
(x
+ y)
2
−
4
(x
− y)
4
2
= xy.
If both odd,
f ( x + y) − f ( x − y) =
(x
+ y) − 1
2
4
−
(x
− y) − 1
2
4
= xy.
Proof by induction that the formula for f ( n) is correct. It is true for n = 1. Assume that the formula is valid for k. If k is even,
then f ( k ) = k 2 4 and
f ( k + 1) = f ( k ) +
(k + 1) − 1.
k
k2
k
k 2 + 2k
=
+
=
=
2
4
2
4
4
2
The argument is similar if k is odd.
Section 9.3 The Integral Test and p-Series
∞
1.
∞
1
∑n + 3
3.
n =1
Let f ( x) =
n =1
1
.
x +3
Let f ( x) =
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
1
∑ 2n
1
.
2x
f is positive, continuous, and decreasing for x ≥ 1.
∞
1
dx = ⎡⎣ln ( x + 3)⎤⎦1 = ∞
x + 3
∞
∫1
Diverges by Theorem 9.10
∞
⎡ −1 ⎤
1
1
=
dx = ⎢
x⎥
2x
ln
2
2
2
ln
2
(
)
⎣⎢
⎦⎥1
Converges by Theorem 9.10.
∞
2
2. ∑
3
n
+5
n =1
Let f ( x) =
∞
4.
n =1
2
.
3x + 5
Let f ( x) =
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
∑ 3− n
1
.
3x
f is positive, continuous, and decreasing for x ≥ 1.
∞
2
⎡2
⎤
dx = ⎢ ln (3 x + 5)⎥ = ∞
3x + 5
⎣3
⎦1
∞
∫1
Diverges by Theorem 9.10
∞
⎡ −1 ⎤
1
1
=
dx = ⎢
x
x⎥
3
3 ln 3
⎢⎣ (ln 3) 3 ⎥⎦1
Converges by Theorem 9.10.
5.
∞
∑ e− n
n =1
Let f ( x ) = e − x .
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
∞
1
e − x dx = ⎣⎡−e − x ⎦⎤ =
1
e
Converges by Theorem 9.10
© 2010 Brooks/Cole, Cengage Learning
288
6.
Chapter 9
Infinite Series
∞
∑ ne−n 2
10.
n =1
n=2
ln n
n
Let f ( x) = xe −x 2 .
Let f ( x) =
f is positive, continuous, and decreasing for x ≥ 3 since
2 − x
f ′( x) =
< 0 for x ≥ 3.
2e x 2
f is positive, continuous, and decreasing for
x > e 2 ≈ 7.4.
∞
∫3
∞
∫2
∞
xe − x 2 dx = ⎡⎣−2( x + 2)e− x 2 ⎤⎦ = 10e−3 2
3
Converges by Theorem 9.10
7.
∞
∑
∞
n =1
ln x
dx = ⎡⎣2
x
11.
∞
∑
n
n =1
1
Let f ( x) = 2
.
x +1
(
1
)
n +1
Let f ( x) =
x
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
1
π
∞
dx = [arctan x]1 =
x +1
4
f ′( x) = −
2
Converges by Theorem 9.10
8.
∞
1
∞
∫1
n =1
1
.
2x + 1
∞
1
dx = ⎡⎣ln
2x + 1
12.
9.
∑
n =1
∫1
)
x +1
2
< 0.
(
1
dx = ⎡2 ln
⎣
x +1
)
(
)
∞
x + 1 ⎤ = ∞, diverges
⎦1
∞
∑ n2
n
+3
x
.
x2 + 3
x ≥ 2 since
n +1
ln ( x + 1)
ln ( x + 1)
x +1
(
x
,
f ( x) is positive, continuous, and decreasing for
x +1
.
∞
⎡ ⎡ln ( x + 1)⎤ 2 ⎤
⎦ ⎥ = ∞
dx = ⎢ ⎣
⎢
⎥
2
⎣
⎦1
Diverges by Theorem 9.10
3 − x2
< 0 for x ≥ 2.
( x 2 + 3)
f ′( x) =
∞
∫1
f is positive, continuous, and decreasing for
1 − ln ( x + 1)
x ≥ 2 because f ′( x) =
< 0 for x ≥ 2.
( x + 1)2
∞
2x
Let f ( x) =
ln ( n + 1)
Let f ( x) =
x
n =1
∞
2 x + 1⎤⎦ = ∞
1
Diverges by Theorem 9.10
∞
1+ 2
32
)
x +1
So, the series diverges by Theorem 9.10.
f is positive, continuous, and decreasing for x ≥ 1.
∫1
(
1
f is positive, continuous, and decreasing for x ≥ 1.
∑ 2n + 1
Let f ( x) =
∞
x (ln x − 2)⎤⎦ = ∞, diverges
2
So, the series diverges by Theorem 9.10.
1
+1
∑ n2
ln x
2 − ln x
, f ′( x) =
.
2 x3 2
x
∞
x
dx = ⎡ln
⎣
x +3
x 2 + 3⎤ = ∞
⎦1
2
Diverges by Theorem 9.10
13.
∞
∑
n =1
1
n + 2
Let f ( x) =
1
−1
< 0
, f ′( x) =
32
x + 2
2( x + 2)
f is positive, continuous, and decreasing for x ≥ 1.
∞
1
∫ 1 ( x + 2)1 2
dx = ⎡⎣2
∞
x + 2 ⎤⎦ = ∞, diverges.
1
So, the series diverges by Theorem 9.10.
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
14.
∞
ln n
3
n=2 n
∑
18.
ln x
1 − 3 ln x
, f ′( x) =
.
x3
x4
f ′( x) =
∞
⎡ ( 2 ln x + 1) ⎤
ln x
dx = ⎢−
⎥
x2
4x4
⎣
⎦2
=
∞
∫3
ln n
15. ∑ 2
n =1 n
19.
ln x
1 − 2 ln x
Let f ( x) = 2 , f ′( x) =
.
x
x3
∞
n = 1 ( 2 n + 3)
⎡−(ln x + 1) ⎤
ln x
dx = ⎢
⎥ = 1, converges
x2
x
⎣
⎦1
∫1
(2 x + 3)
So,
∑
∞
20.
∞
−3
⎡
⎤
1
−1
⎥ =
dx = ⎢
, converges.
2
100
⎣⎢ 4( 2 x + 3) ⎦⎥1
1
( 2n
+ 3)
3
converges by Theorem 9.10.
n + 2
∑n +1
Let f ( x) =
∞
1
dx = ⎡⎣2 ln x ⎤⎦ = ∞, diverges
2
x ln x
−1
x+2
1
< 0
= 1+
, f ′( x) =
2
x +1
x +1
x
( + 1)
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
So, the series diverges by Theorem 9.10.
∞
∞
x + 2
dx = ⎡⎣ x + ln ( x + 1)⎤⎦1 diverges.
x +1
arctan n
2
n =1 n + 1
So,
arctan x
Let f ( x) = 2
,
x +1
[Note: lim
∑
∞
n + 2
∑ n + 1 diverges by Theorem 9.10.
n =1
1 − 2 x arctan x
( x 2 + 1)
2
n→∞
< 0 for x ≥ 1.
21.
∞
4n
+1
n =1
2 ∞
⎡ (arctan x) ⎤
arctan x
3π 2
⎥ =
dx = ⎢
, converges
2
x +1
2
32
⎢⎣
⎥⎦1
So, the series converges by Theorem 9.10.
n + 2
= 1 ≠ 0, so the series diverges.]
n +1
∑ 2n 2
f is positive, continuous, and decreasing for x ≥ 1.
∞
< 0
n =1
f is positive, continuous, and decreasing for x ≥ 2.
f ′( x) =
4
∞
∞
n =1
1
2 ln x + 1
Let f ( x) =
, f ′( x) = −
.
32
x ln x
2 x 2 (ln x)
−6
(2 x + 3)
f is positive, continuous, and decreasing for x ≥ 1.
∞
1
16. ∑
n = 2 n ln n
∫1
3
−3
∞
17.
1
∑
So, the series converges by Theorem 9.10.
∞
∞
1
dx = ⎡⎣ln (ln (ln x))⎤⎦ = ∞, diverges
3
x ln x ln (ln x)
Let f ( x) = ( 2 x + 3) , f ′( x) =
f is positive, continuous, and decreasing for
x > e1 2 ≈ 1.6.
∫2
−⎡⎣(ln x + 1) ln (ln x) + 1⎤⎦
.
2
2
x 2 (ln x) (ln (ln x))
So, the series diverges by Theorem 9.10.
∞
∞
1
,
x ln x ln (ln x)
f is positive, continuous, and decreasing for x ≥ 3.
2 ln 2 + 1
, converges
16
So, the series converges by Theorem 9.10.
∫1
1
∑ n ln n ln(ln n)
Let f ( x) =
f is positive, continuous, and decreasing for x > 2.
∞
∞
289
n=3
Let f ( x) =
∫2
The Integral Test and p-Series
Let f ( x) =
−4( 2 x 2 − 1)
4x
′
,
f
x
=
< 0
(
)
2
2 x2 + 1
(2 x 2 + 1)
for x ≥ 1
f is positive, continuous, and decreasing for x ≥ 1.
∞
∫1
So,
∞
4x
dx = ⎡⎣ln ( 2 x 2 + 1)⎤⎦ = ∞, diverges.
1
2x2 + 1
∞
4n
diverges by Theorem 9.10.
+1
∑ 2n 2
n =1
© 2010 Brooks/Cole, Cengage Learning
290
22.
Chapter 9
∞
∑ n4
n =1
Infinite Series
n
+1
26.
n =1
x
1 − 3x
, f ′( x) =
< 0 for x > 1.
2
x +1
( x 4 + 1)
4
Let f ( x) =
Let f ( x) =
4
∞
∞
x
π
⎡1
⎤
dx = ⎢ arctan ( x 2 )⎥ = , converges
x +1
2
8
⎣
⎦1
f ′( x) =
4
So, the series converges by Theorem 9.10.
23.
∞
n = 1 ( 4 n + 5)
∞
∫1
32
x
Let f ( x) =
(4 x
+ 5)
32
−( 2 x − 5)
, f ′( x) =
(4 x
+ 5)
52
∫ 1 (4 x + 5)3 2
(Integration by parts: u = x, dv = ( 4 x + 5)
∞
∑
So,
n =1
24.
∞
n
( 4n
+ 5)
32
n =1
Let f ( x) =
∞
∑
n =1
x
(x
2
−3 2
+ 1)
2
+ 1)
, f ′( x) =
∫1
( x2
2
n
∑ nk
( x2
+ 1)
3
< 0 for
⎡
1
−1 ⎤
⎥ = , converges
dx = ⎢ 2
4
⎢⎣ 2( x + 1) ⎥⎦
1
∞
31.
x >
k
Let f ( x) =
∫1
f ′( x) =
for x >
x
k
k −2
∞
1
1
⎡ 1 ⎤
dx = ⎢− 2 ⎥ =
x3
2
x
2
⎣
⎦1
Converges by Theorem 9.10
32.
∞
1
∑ n1 3
n =1
⎡⎣c( k − 1) − x ⎤⎦
< 0
2
( x k + c)
k
c( k − 1) .
Let f ( x) =
x k −1
⎡1
⎤
dx = ⎢ ln ( x k + c)⎥ = ∞
x + c
⎣k
⎦1
1
.
x1 3
f is positive, continuous, and decreasing for x ≥ 1.
∞
∞
k
1
.
x3
f is positive, continuous, and decreasing for x ≥ 1.
k −1
c( k − 1) because
1
∑ n3
n =1
+ c
x
.
xk + c
f is positive, continuous, and decreasing for
∞
The function f is not decreasing for x ≥ 1.
∞
Let f ( x) =
∫1
2
⎛ sin x ⎞
30. Let f ( x ) = ⎜
⎟ , f ( n ) = an .
⎝ x ⎠
k −1
n =1
2 + sin x
, f ( n ) = an .
x
The function f is not decreasing for x ≥ 1.
n
∑ n4 + 2n2 + 1 converges by Theorem 9.10.
n =1
∞
f ( n ) = an .
The function f is not positive for x ≥ 1.
∞
So,
25.
+ 1)
x
28. Let f ( x ) = e− x cos x, f ( n) = an .
∞
x
(−1) x ,
2
x ≥1
f is positive, continuous, and decreasing for x ≥ 1
∞
k!
e
Converges by Theorem 9.10
29. Let f ( x ) =
2
−(3 x − 1)
+
The function f is not positive for x ≥ 1.
dx )
n
(n2
∞
k ( k − 1)
1
k
+
+
+
e
e
e
=
diverges by Theorem 9.10.
n
=
+ 2n 2 + 1
∑ n4
∞
27. Let f ( x) =
∞
⎡ 2x + 5 ⎤
= ⎢
⎥ = ∞
⎣ 4 4 x + 5 ⎦1
x
< 0 for x > k .
ex
x k e − x dx = ⎡⎣− x k e − x ⎤⎦ + k ∫ x k −1e − x dx
1
1
< 0 for
x ≥ 3
f is positive, continuous, and decreasing for x ≥ 3.
∞
x k −1 ( k − x)
Use integration by parts.
n
∑
xk
.
ex
f is positive, continuous, and decreasing for
x > k because
f is positive, continuous, and decreasing for x > 1.
∫1
∞
∑ nk e− n
∫1
∞
1
⎡3
⎤
dx = ⎢ x 2 3 ⎥ = ∞
x1 3
⎣2
⎦1
Diverges by Theorem 9.10
Diverges by Theorem 9.10
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
33. Let f ( x) =
1
−1
, f ′( x) =
< 0 for x ≥ 1
14
x
4 x5 4
42.
∞
∞
43.
S1 = 2
1
−4
, f ′( x) = 5 < 0 for x ≥ 1
x4
x
S 2 ≈ 3.1892
S3 ≈ 4.0666
f is positive, continuous, and decreasing for x ≥ 1
∞
∫1
35.
∑5
n =1
1
=
n
∞
∞
44.
S1 = 2
1
S3 ≈ 3.6667
n =1
S 4 ≈ 4.1667
1
<1
5
3
∑ n5 3
Matches (f), diverges
45.
5
>1
3
Note: The partial sums for 45 and 47 are very similar
because π 2 ≈ 3 2.
46.
n =1
2
S1 = 2
S 2 ≈ 3.5157
1
∑ n3 2
S3 ≈ 4.8045
n =1
Convergent p-series with p =
Matches (a), diverges
3
>1
2
47.
1
∞
∑n
n =1
∑ n2 3
2
n
=
∞
2
∑ n3 2
n =1
S1 = 2
n =1
∞
∞
∑ n2 5
n =1
Convergent p-series with p = 2 > 1
41.
n =1
Matches (b), converges
1
Divergent p-series with p =
2
S 4 ≈ 3.2560
∑ n2
∞
∞
∑ nπ 2
S3 ≈ 3.0293
1
Divergent p-series with p =
<1
2
40.
nπ
=
S 2 ≈ 2.6732
1
37. ∑ 1 2
n =1 n
39.
2
S1 = 2
∞
∞
∞
∑
n =1
Convergent p-series with p =
∞
2
S2 = 3
n =1
38.
∞
∑n
n =1
∑ n1 5
Divergent p-series with p =
36.
Matches (c), diverges
1
1
⎡ −1 ⎤
dx = ⎢ 3 ⎥ = , converges
x4
3
x
3
⎣
⎦1
1
So, ∑ 4 converges by Theorem 9.10.
n
n =1
∞
S 4 ≈ 4.7740
∞
∞
2
∑ n3 4
n =1
So, the series diverges by Theorem 9.10.
34. Let f ( x) =
1
∑ nπ
Convergent p-series with p = π > 1
34 ∞
⎡4x ⎤
1
dx = ⎢
⎥ = ∞, diverges
x1 4
⎣ 3 ⎦1
291
n =1
f is positive, continuous, and decreasing for x ≥ 1
∫1
∞
The Integral Test and p-Series
2
<1
3
1
∑ n1.04
n =1
Convergent p-series with p = 1.04 > 1
S 2 ≈ 2.7071
S3 ≈ 3.0920
S 4 ≈ 3.3420
Matches (d), converges
Note: The partial sums for 45 and 47 are very similar
because π 2 ≈ 3 2.
© 2010 Brooks/Cole, Cengage Learning
292
48.
Chapter 9
∞
Infinite Series
2
∑ n2
n =1
S1 = 2
S 2 = 2.5
S3 ≈ 2.7222
Matches (e), converges
49. (a)
n
5
10
20
50
100
Sn
3.7488
3.75
3.75
3.75
3.75
The partial sums approach the sum 3.75 very rapidly.
11
0
11
0
(b)
n
5
10
20
50
100
Sn
1.4636
1.5498
1.5962
1.6251
1.635
The partial sums approach the sum π 2 6 ≈ 1.6449 slower than the series in part (a).
8
0
12
0
N
50.
1
∑n
1 1 1
+ + +
2 3 4
=1+
n =1
(a)
+
1
> M
N
M
2
4
6
8
N
4
31
227
1674
51. Let f be positive, continuous, and decreasing for
x ≥ 1 and an = f ( n). Then,
∑ an
n =1
and
∞
1
1
1
=
+
+
n
10,000
10,001
n = 10,000
∑
is the harmonic series, starting with the 10,000th term,
and therefore diverges.
(b) No. Because the terms are decreasing (approaching
zero), more and more terms are required to increase
the partial sum by 2.
∞
53. Your friend is not correct. The series
54. No. Theorem 9.9 says that if the series converges, then
the terms an tend to zero. Some of the series in Exercises
43–48 converge because the terms tend to 0 very rapidly.
6
55.
∑ an
n =1
∫ 1 f ( x) dx
either both converge or both diverge (Theorem 9.10).
See Example 1, page 620.
∞
≥
7
∑ an
n=2
y
∞
52. A series of the form
7
∫1 f ( x) dx
≥
1
∑ n p is a p-series,
p > 0.
1
x
1
2
3
4
5
6
7
n =1
The p-series converges if p > 1 and diverges if
0 < p ≤ 1.
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
56. (a)
The Integral Test and p-Series
293
y
1
x
1
∞
2
3
1
>
n
∑
n =1
4
1
dx
x
∞
∫1
The area under the rectangle is greater than the area under the curve.
(b)
1
dx = ⎡⎣2
x
∞
∫1
Because
x ⎤⎦
∞
= ∞, diverges,
1
∞
∑
n =1
1
diverges.
n
y
1
x
1
∞
2
1
∑ n2
3
∞
∫1
<
n=2
4
1
dx
x2
The area under the rectangles is less than the area under the curve.
Because
∞
1
∑ n2
n=2
57.
∞
∞
1
⎡ 1⎤
dx = ⎢− ⎥ = 1, converges,
x2
⎣ x ⎦1
∞
∫1
⎛
converges ⎜ and so does
⎝
∞
1⎞
∑ n2 ⎟.
n =1
⎠
1
∑ n ln n p
n=2
(
)
If p = 1, then the series diverges by the Integral Test. If p ≠ 1,
∞
1
∫ 2 x(ln x) p
∞
dx =
∞
∫ 2 (ln x)
−p
⎡ (ln x)− p + 1 ⎤
1
⎥ .
dx = ⎢
x
⎢⎣ − p + 1 ⎥⎦ 2
Converges for − p + 1 < 0 or p > 1
58.
∞
∑
n=2
ln n
np
If p = 1, then the series diverges by the Integral Test. If p ≠ 1,
∞
∫2
ln x
dx =
xp
∞
∞
∫2
x
−p
⎡ x − p +1
⎤
ln x dx = ⎢
⎡−1 + (− p + 1) ln x⎤⎦⎥ . (Use integration by parts.)
2⎣
⎢⎣ ( − p + 1)
⎥⎦ 2
Converges for − p + 1 < 0 or p > 1
© 2010 Brooks/Cole, Cengage Learning
294
59.
Chapter 9
∞
∑
n =1
Infinite Series
n
(1 + n2 )
64.
p
∞
n
If p = 1, ∑
diverges (see Example 1). Let
2
n =1 1 + n
x
f ( x) =
(1 + x )
2
p
,p ≠1
(
So,
∑
(1 + x 2 )
p +1
.
∞
(1 + x 2 )
For p > 1, this integral converges. For 0 < p < 1, it
diverges.
(
∞
∑ n(1 + n2 )
1
(
n=2
p > 1.
)
∞
1
∞
So,
(
)
1
(
p > 1.
)
∞
1
∑ n ln n2
∞
1
∑ 2n ln n
=
( )
n=2
67.
)
∑ n ln n 2 , converges.
=
n=2
1 ∞
1
, diverges
∑
2 n = 2 n ln ( n)
y
f(1) = a1
f(2) = a2
1
60.
2
1
n=2
⎡
⎤
1
⎥
dx = ⎢
p
1
−
⎢ x2 + 1
(⎣
) (2 − 2 p) ⎥⎦1
p
n 3 (ln n)
∞
∑ n ln n 2 3 , diverges.
=
∑ n ln n p converges for
∞
∫1
1
n=2
is positive, continuous, and eventually decreasing.
x
∞
∑ n ln n p converges for
So,
66.
p > 1.
)
n=2
65.
For a fixed p > 0, p ≠ 1, f ′( x) is eventually negative. f
∞
1
n=2
n=2
1 − ( 2 p − 1) x 2
f ′( x) =
∞
∑ n ln n p converges for
f(N + 1) = aN + 1
f(N) = aN
p
n =1
Because p > 0, the series diverges for all values of p.
61.
∞
∑
n =1
1
=
pn
∞
n
⎛1⎞
∑ ⎜ p ⎟ , Geometric series.
n =1 ⎝ ⎠
SN =
62.
∑
n=3
1
p
⎡ ⎡ln (ln x)⎤ − p + 1 ⎤
⎦
⎢⎣
⎥ .
dx
=
p
⎢
p
1 ⎥
−
+
x ln x ⎡⎣ln (ln x)⎤⎦
⎣
⎦3
1
This converges for − p + 1 < 0 ⇒ p > 1.
So, the series converges for p > 1, and diverges for
0 < p ≤ 1.
∑ n ln n p
n=2
(
∞
)
∞
∑
≤
So, 0 ≤ Rn ≤
an > 0
∞
∑
an = aN +1 + aN + 2 +
converges for p > 1.
1
, diverges.
So, ∑
ln
n
n
n=2
∞
∫ N f ( x) dx
∞
∫ N f ( x) dx
68. From Exercise 67, you have:
∞
∫ N f ( x) dx
0 ≤ S − SN ≤
1
+ aN
n = N +1
n ln n ⎣⎡ln (ln n)⎦⎤
∞
= a1 + a2 +
RN = S − S N =
∞
63.
N
∑ an
n = N +1
If p ≠ 1,
∫3
N N+1
RN = S − S N =
If p = 1, then
∞
∞
1
∫ 3 x ln x ⎡ln(ln x)⎤ dx = ⎡⎣ln(ln(ln x))⎤⎦3 = ∞, so the
⎣
⎦
series diverges by the Integral Test
∞
x
...
2
n =1
1
<1⇒ p >1
Converges for
p
∞
1
SN ≤ S ≤ SN +
N
∑ an
≤ S ≤
n =1
69. S6 = 1 +
R6 ≤
∞
∫6
1.0811 ≤
N
∞
∫ N f ( x) dx
∞
∑ an + ∫ N f ( x) dx
n =1
1
1
1
1
1
+ 4 + 4 + 4 + 4 ≈ 1.0811
24
3
4
5
6
∞
1
⎡ 1 ⎤
dx = ⎢− 3 ⎥ ≈ 0.0015
x4
⎣ 3x ⎦ 6
∞
1
∑ n4
≤ 1.0811 + 0.0015 = 1.0826
n =1
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
295
1
1
1
+ 5 + 5 ≈ 1.0363
5
2
3
4
70. S 4 = 1 +
R4 ≤
The Integral Test and p-Series
∞
1
⎡ 1 ⎤
dx = ⎢− 4 ⎥ ≈ 0.0010
5
x
⎣ 4x ⎦4
∞
∫4
∞
1
∑ n5
1.0363 ≤
≤ 1.0363 + 0.0010 = 1.0373
n =1
1
1
1
1
1
1
1
1
1
1
+ +
+
+
+
+
+
+
+
≈ 0.9818
2 5 10 17
26 37
50
65 82 101
∞
1
π
∞
R10 ≤ ∫
dx = [arctan x]10 =
− arctan 10 ≈ 0.0997
10 x 2 + 1
2
∞
1
0.9818 ≤ ∑ 2
≤ 0.9818 + 0.0997 = 1.0815
n
+1
n =1
71. S10 =
72. S10 =
1
2(ln 2)
1
+
3
3(ln 3)
3
+
1
4(ln 4)
+
3
+
1
11(ln 11)
3
≈ 1.9821
∞
R10
⎡
⎤
1
1
⎢−
⎥ =
≤ ∫
dx
=
≈ 0.0870
3
2
3
10
⎢
⎥
2(ln 11)
( x + 1)⎡⎣ln ( x + 1)⎤⎦
⎣ 2 ⎡⎣ln ( x + 1)⎤⎦ ⎦10
1
∞
1.9821 ≤
∞
∑
n =1
73. S 4 =
R4 ≤
1
(n
+ 1) ⎡⎣ln ( n + 1)⎤⎦
3
≤ 1.9821 + 0.0870 = 2.0691
1
2
3
4
+ 4 + 9 + 16 ≈ 0.4049
e
e
e
e
∞
∫4
∞
−16
2
e
⎡ 1 2⎤
xe − x dx = ⎢− e − x ⎥ =
2
⎣ 2
⎦4
∞
∑ ne− n
0.4049 ≤
2
≈ 5.6 × 10−8
≤ 0.4049 + 5.6 × 10−8
R4 ≤
∞
∞
0.5713 ≤
∑ e− n
N ≥ 2
≤ 0.5713 + 0.0183 = 0.5896
n=0
75. 0 ≤ RN ≤
∞
∫N
∞
1
1
⎡ 1 ⎤
dx = ⎢− 3 ⎥ =
< 0.001
3N 3
x4
⎣ 3x ⎦ N
1
< 0.003
N3
78. RN ≤
∞
∫N e
−x 2
∞
2
dx = ⎡⎣−2e − x 2 ⎤⎦ = N 2 < 0.001
N
e
2
< 0.001
eN 2
eN
2
> 2000
N
> ln 2000
2
N > 2 ln 2000 ≈ 15.2
N 3 > 333.33
N > 6.93
N ≥ 16
N ≥ 7
76. 0 ≤ RN ≤
e5 N > 200
N >
e − x dx = ⎡⎣−e − x ⎤⎦ ≈ 0.0183
4
∞
1
< 0.005
e5 N
ln 200
5
N > 1.0597
1
1
1
1
+ 2 + 3 + 4 ≈ 0.5713
e
e
e
e
∫4
∫N
∞
e −5 N
⎡ 1
⎤
e −5 x dx = ⎢− e −5 x ⎥ =
< 0.001
5
⎣ 5
⎦N
5 N > ln 200
n =1
74. S 4 =
∞
77. RN ≤
∞
∫N
∞
1
⎡ 2 ⎤
dx = ⎢− 1 2 ⎥ =
x3 2
⎣ x ⎦N
2
< 0.001
N
N −1 2 < 0.0005
N > 2000
N ≥ 4,000,000
© 2010 Brooks/Cole, Cengage Learning
296
Chapter 9
∞
∫N
79. RN ≤
Infinite Series
1
∞
dx = [arctan x]N
x +1
80. Rn ≤
2
=
−arctan N < 0.001 −
π
arctan N >
2
π
2
∞
∫N
∞
⎡ 1
2
⎛ x ⎞⎤
dx = 2 ⎢
arctan ⎜
⎟⎥
2
x +5
⎝ 5 ⎠⎦ N
⎣ 5
− arctan N < 0.001
=
π
2
π
− 0.001
2
2 ⎛π
⎛ N ⎞⎞
⎜ − arctan ⎜
⎟ ⎟ < 0.001
5⎝ 2
⎝ 5 ⎠⎠
⎛ N ⎞
− arctan ⎜
⎟ < 0.0005 5
⎝ 5⎠
π
⎛ N ⎞
−arctan ⎜
⎟ < 0.0005 5 −
2
⎝ 5⎠
⎛π
⎞
N > tan ⎜ − 0.001⎟
⎝2
⎠
N ≥ 1000
π
⎛ N ⎞
arctan ⎜
− 0.0005 5
⎟ >
2
5
⎝
⎠
⎛π
⎞
5 tan ⎜ − 0.0005 5 ⎟
⎝2
⎠
N ≥ 1000
N >
81. (a)
∞
1
∑ n1.1 . This is a convergent p-series with
n=2
f ( x) =
∞
∫2
(b)
6
1
is a divergent series. Use the Integral Test.
n
n
ln
n=2
1
is positive, continuous, and decreasing for x ≥ 2.
x ln x
∞
1
dx = ⎣⎡ln ln x ⎤⎦ 2 = ∞
x ln x
1
∑ n1.1
1
1
1
1
1
+ 1.1 + 1.1 + 1.1 + 1.1 ≈ 0.4665 + 0.2987 + 0.2176 + 0.1703 + 0.1393
21.1
3
4
5
6
=
n=2
6
∞
p = 1.1 > 1. ∑
1
∑ n ln n
=
n=2
1
1
1
1
1
+
+
+
+
≈ 0.7213 + 0.3034 + 0.1803 + 0.1243 + 0.0930
2 ln 2
3 ln 3
4 ln 4
5 ln 5
6 ln 6
For n ≥ 4, the terms of the convergent series seem to be larger than those of the divergent series.
(c)
1
1
<
1.1
n
n ln n
n ln n < n1.1
ln n < n 0.1
This inequality holds when n ≥ 3.5 × 1015. Or, n > e 40 . Then ln e40 = 40 < (e 40 )
82. (a)
∞
∫ 10
∞
⎡ x − p +1 ⎤
1
dx = ⎢
⎥ =
p
x
⎣ − p + 1⎦10
(b) f ( x) =
(p
0.1
= e 4 ≈ 55.
1
,p >1
− 1)10 p −1
1
xp
R10 ( p ) =
∞
1
p
n
n = 11
∑
≤ Area under the graph of f over the interval [10, ∞)
(c) The horizontal asymptote is y = 0. As n increases, the error decreases.
© 2010 Brooks/Cole, Cengage Learning
Section 9.3
The Integral Test and p-Series
297
83. (a) Let f ( x) = 1 x. f is positive, continuous, and decreasing on [1, ∞).
1
dx
x
S n − 1 ≤ ln n
n
∫1
Sn − 1 ≤
So, S n ≤ 1 + ln n. Similarly,
n +1
∫1
Sn ≥
1
dx = ln ( n + 1).
x
So, ln ( n + 1) ≤ S n ≤ 1 + ln n.
y
1
1
2
1
x
3 ...
n −1
2
n
n+1
(b) Because ln ( n + 1) ≤ Sn ≤ 1 + ln n, you have ln ( n + 1) − ln n ≤ Sn − ln n ≤ 1. Also, because ln x is an increasing
function, ln ( n + 1) − ln n > 0 for n ≥ 1. So, 0 ≤ S n − ln n ≤ 1 and the sequence {an} is bounded.
(c) an − an + 1 = [S n − ln n] − ⎡⎣S n + 1 − ln ( n + 1)⎤⎦ =
n +1
∫n
1
1
dx −
≥ 0
x
n +1
So, an ≥ an + 1 and the sequence is decreasing.
(d) Because the sequence is bounded and monotonic, it converges to a limit, γ .
(e) a100 = S100 − ln 100 ≈ 0.5822 ( Actually γ ≈ 0.577216.)
84.
∞
⎛
∑ ln⎜⎝1 −
n=2
⎛ n2 − 1⎞
⎟ =
2
⎝ n ⎠
∞
1⎞
⎟ =
n2 ⎠
∑ ln⎜
n=2
(
∞
∑ ln
(n
+ 1)( n − 1)
n
n=2
2
) (
=
∞
∑ ⎡⎣ln(n + 1) + ln(n − 1) − 2 ln n⎤⎦
n=2
) (
) (
= ln 3 + ln 1 − 2 ln 2 + ln 4 + ln 2 − 2 ln 3 + ln 5 + ln 3 − 2 ln 4 + ln 6 + ln 4 − 2 ln 5
(
) (
) (
)
+ ln 7 + ln 5 − 2 ln 6 + ln 8 + ln 6 − 2 ln 7 + ln 9 + ln 7 − 2 ln 8 +
85.
= −ln 2
∞
∑ xln n
n=2
(a) x = 1:
∞
∑ 1ln n
=
n=2
(b) x =
1
:
e
∞
∞
∑ 1, diverges
n=2
⎛1⎞
∑ ⎜⎝ e ⎟⎠
ln n
=
n=2
∞
∑ e−ln n
∞
n=2
n=2
(c) Let x be given, x > 0. Put x = e
∞
∑ xln n
=
n=2
∞
∑ e− p ln n
n=2
=
∞
∑ n− p
1
∑ n , diverges
=
−p
=
n=2
⇔ ln x = − p.
∞
1
∑ np
n=2
This series converges for p > 1 ⇒ x <
86. ξ ( x) =
∞
∑ n− x
n =1
=
∞
1
.
e
1
∑ nx
n =1
Converges for x > 1 by Theorem 9.11
© 2010 Brooks/Cole, Cengage Learning
)
298
Chapter 9
Infinite Series
1
−3
, f ′( x) =
< 0 for x ≥ 1
3x − 2
( 3 x − 2) 2
87. Let f ( x) =
94.
∞
∞
So, the series
95.
1
∑ 3n − 2
1
n=2
n n2 − 1
x2 − 1
x
1
∞
dx = [arcsec x]2 =
∞
x2 − 1
1
∑ n4
n =1
n
π
n =1
n
∑ ln(n)
lim ln ( n) = ∞
2
−
Diverges by Theorem 9.9
3
97.
∞
1
∑ n ln n 3
n=2
1
∑ n5 4
n =1
(
)
Let f ( x) =
5
4
1
x(ln x)
∞
1
∫ 2 x(ln x)3 dx
1
n 0.95
∞
∞
∑ (1.042)
n
is geometric with r = 1.042 > 1. Diverges
n=0
n =1
lim
n→∞
dx
∞
98.
n
n +1
2
n
n2 + 1
ln n
3
n=2 n
∑
Let f ( x) =
ln x
.
x3
f is positive, continuous, and decreasing for x ≥ 2 since
1 − 3 ln x
f ′( x) =
< 0 for x ≥ 2.
x4
by Theorem 9.6.
∞
x
Converges by Theorem 9.10. See Exercise 57.
2
3
Converges by Theorem 9.6
∑
−3 1
∞
n
Geometric series with r =
93.
∞
∫ 2 (ln x)
⎡
1 ⎤
1
= ⎢−
⎥ =
2
2
2(ln 2)
⎢⎣ 2(ln x) ⎥⎦ 2
n=0
92.
=
∞
Diverges by Theorem 9.11
∑ ( 23 )
.
⎡ (ln x)−2 ⎤
= ⎢
⎥
⎢⎣ −2 ⎥⎦ 2
p-series with p = 0.95
91.
3
f is positive, continuous, and decreasing for x ≥ 2.
Converges by Theorem 9.11
∞
1⎞
⎛
∑ ⎜⎝1 + n ⎟⎠
n =1
n→∞
π
∞
=
p-series with p =
90. 3∑
∞
n=2
Converges by Theorem 9.10
89.
1
n =1
∞
96.
.
f is positive, continuous, and decreasing for x ≥ 2.
x
n =1
∞
∑ n3
Diverges by Theorem 9.9
1
Let f ( x) =
∫2
−
Fails nth-Term Test
∞
∞
1
n
diverges by Theorem 9.10.
∑
∞
∑ n2
1⎞
⎛
lim ⎜1 + ⎟ = e ≠ 0
n→∞ ⎝
n⎠
n =1
88.
1⎞
⎟ =
n3 ⎠
−
Because these are both convergent p-series, the
difference is convergent.
∞
1
⎡1
⎤
dx = ⎢ ln 3 x − 2 ⎥ = ∞
3x − 2
⎣3
⎦1
⎛1
n =1
f is positive, continuous, and decreasing for x ≥ 1.
∫1
∞
∑ ⎜⎝ n2
∞
= lim
n→∞
1
1 + (1 n 2 )
=1≠ 0
∫2
∞
ln x
1 ∞ 1
⎡ ln x ⎤
dx = ⎢− 2 ⎥ + ∫
dx
3
2 2 x3
x
⎣ 2x ⎦2
∞
=
ln 2 ⎡ 1 ⎤
+ ⎢− 2 ⎥
8
⎣ 4x ⎦2
=
ln 2
1
+
( Use integration by parts.)
8
16
Diverges by Theorem 9.9
Converges by Theorem 9.10. See Exercise 36.
© 2010 Brooks/Cole, Cengage Learning
Section 9.4
Comparisons of Series
299
Section 9.4 Comparisons of Series
∞
6
∑ n3 2
1. (a)
6
6
+ 32 +
1
2
=
n =1
∞
∑ n3 2
n =1
; S1 = 6
6
6
6
=
+ 32
+
4
2 +3
+3
∞
an
an = 6
n 3/2
6
; S1 =
6
6
6
=
+
+
∑
2
1 1.5
2 4.5
n + 0.5
n =1 n
3
2
; S1 =
5
an =
4
n
6
n 2 + 0.5
3
6
≈ 4.9
1.5
an =
2
6
n 3/2 + 3
1
n
2
3
⎛
⎞
> 1⎟.
(b) The first series is a p-series. It converges ⎜ p =
2
⎝
⎠
4
6
8
10
(c) The magnitude of the terms of the other two series are less than the corresponding terms at the convergent p-series. So, the
other two series converge.
(d) The smaller the magnitude of the terms, the smaller the magnitude of the terms of the sequence of partial sums.
Sn
n
Σ
k=1
12
6
k 3/2
n
Σ
10
k=1
k
6
k 2 + 0.5
8
6
4
n
Σ
2
k=1
6
k 3/2 + 3
n
2
∞
6
8
2
= 2+
n
∑
2. (a)
4
n =1
∞
10
2
+
2
2
2
=
+
0.5
n − 0.5
∑
n =1
∞
4
=
n + 0.5
∑
n =1
4
+
1.5
an
S1 = 2
2
n − 0.5
an =
4
2
+
2 − 0.5
4
+
2.5
S1 = 4
an =
3
4
n + 0.5
an = 2
n
2
S1 ≈ 3.3
1
n
2
1
⎛
⎞
< 1⎟.
(b) The first series is a p-series. It diverges ⎜ p =
2
⎝
⎠
4
6
8
10
(c) The magnitude of the terms of the other two series are greater than the corresponding terms of the divergent p-series. So,
the other two series diverge.
(d) The larger the magnitude of the terms, the larger the magnitude of the terms of the sequence of partial sums.
Sn
Σ
20
16
Σ
2
n − 0.5
4
n + 0.5
12
8
4
Σ
2
n
8
10
n
2
3. 0 <
4
6
1
1
< 2
n2 + 1
n
Therefore,
∞
∑ n2
n =1
1
+1
converges by comparison with the convergent p-series
∞
1
∑ n2 .
n =1
4.
1
1
<
3n 2 + 2
3n 2
Therefore,
∞
∑ 3n2
n =1
1
+ 2
converges by comparison with the convergent p-series
1∞ 1
∑ .
3 n =1 n 2
© 2010 Brooks/Cole, Cengage Learning
300
5.
Chapter 9
Infinite Series
1
1
>
> 0 for n ≥ 1
2n − 1
2n
Therefore,
1
∑ 2n − 1
n =1
∞
1 ∞ 1
∑ .
2 n =1 n
1
for n ≥ 2
n
Therefore,
converges by comparison with the convergent p-series
∞
n =1
11. For n > 3,
∞
n=2
7.
∞
n=0
converges by comparison with the convergent p-series
∞
n =1
1
1
< n
4 +1
4
12.
Therefore,
n=0
1
+1
∞
n =1
diverges by comparison with the divergent p-series
1
1 ∞ 1
∑ .
4 n =1 4 n
n=0
4n
⎛ 4⎞
< ⎜ ⎟
n
5 +3
⎝5⎠
n
13. 0 <
Therefore,
∞
∑ 5n
n=0
4n
+3
⎛ 4⎞
∞
∑
n=0
n
1
en
en
2
n
⎛1⎞
∑ ⎜⎝ e ⎟⎠ .
n=0
ln n
1
>
> 0.
n +1
n +1
Therefore,
n
14.
3n
⎛ 3⎞
> ⎜ ⎟ for n ≥ 1
n
2 − 1 ⎝ 2⎠
Therefore,
∞
ln n
∑n +1
n =1
∞
diverges by comparison with the divergent series
∞
1
∑ n + 1.
n =1
Note:
≤
1
∞
n=0
∞
e
n2
converges by comparison with the convergent geometric
series
∑ ⎜⎝ 5 ⎟⎠ .
9. For n ≥ 3,
1
Therefore,
converges by comparison with the convergent geometric
series
∞
1
n −1
∑ 43
∑ 4n .
8.
1
1
> 4
43 n − 1
4 n
Therefore,
converges by comparison with the convergent geometric
series
∞
1
∑ n2 .
n
∞
1
∑ n!
1
.
n
∑ 4n
1
1
>
> 0.
n2
n!
Therefore,
diverges by comparison with the divergent p-series
∑
1
∑ n3 2 .
1
n −1
n=2
1
n3 + 1
n =1
1
>
n −1
∞
n +1
∑
diverges by comparison with the divergent p-series
∑
1
n3 2
<
3
Therefore,
∞
6.
1
10.
n =1
diverges by comparison with the divergent geometric
series
∞
1
∑ n + 1 diverges by the Integral Test.
3n
−1
∑ 2n
⎛ 3⎞
n
∑ ⎜⎝ 2 ⎟⎠ .
n =1
n =1
© 2010 Brooks/Cole, Cengage Learning
Section 9.4
n ( n 2 + 1)
15. lim
1n
n→∞
= lim
n→∞
n2
=1
n2 + 1
(n3 − 2n + 3)
(n + 5)
20. lim
Comparisons of Series
1 n2
n →∞
Therefore,
∞
∑ n2
n =1
∞
∞
1
∑ n.
n =1
1 4n
n→∞
n =1
∞
5 ⋅ 4n
= lim n
= 5
n→∞ 4 + 1
Therefore,
5
∑ 4n + 1
n =1
n
n +1
2
=1
1
∑
∞
(
∞
n( n 2 + 4)
=1
+ 4)
1
n =1
1 n 2 ( n + 3)
1n
3
= lim
n→∞
n3
=1
n ( n + 3)
2
Therefore,
∞
1
1
∑ n2 (n + 3)
∑ n.
n =1
n =1
(2n
+ 1) (5n + 1)
(2 5)n
n→∞
2 n + 1 5n
= lim n
⋅ n =1
n→∞ 5 + 1
2
Therefore,
converges by a limit comparison with the convergent
p-series
∞
1
∑ n3 .
n =1
∞
2n + 1
∑ 5n + 1
n =1
converges by a limit comparison with the convergent
geometric series
n
∞
+ 3)n 2
∑ n2 .
n→∞
n2 + 1
18. lim
n→∞
(n
n +3
∑ n n2
22. lim
diverges by a limit comparison with the divergent p-series
∞
= lim
Therefore,
Therefore,
∞
1n
2
converges by a limit comparison with the convergent
p-series
n2 + 1
= lim
n→∞
1n
n→∞
+ 3) n( n 2 + 4)
(n
n =1
n
⎛1⎞
∑ ⎜⎝ 4 ⎟⎠ .
n =1
1
n =1
n→∞
converges by a limit comparison with the convergent
geometric series
∞
1
∑ n2 .
21. lim
∞
n=0
n +5
− 2n + 3
∑ n3
converges by a limit comparison with the convergent
p-series
5 ( 4n + 1)
17. lim
⎛ n + 5 ⎞⎛ n 2 ⎞
= lim ⎜ 3
⎟⎜ ⎟
n →∞ ⎝ n − 2 n + 3 ⎠ 1
⎝ ⎠
=1
Therefore,
n
+1
diverges by a limit comparison with the divergent p-series
16. lim
301
⎛ 2⎞
∑⎜ ⎟ .
n =1 ⎝ 5 ⎠
2n 2 − 1
2n5 − n3
2
=
19. lim 3n + 23n + 1 = lim 5
n→∞
n → ∞ 3n + 2 n + 1
1n
3
5
Therefore,
23. lim
(
1 n n2 + 1
1n
n→∞
2
)=
lim
n→∞
n2
n n2 + 1
=1
Therefore,
∞
1
n =1
n n2 + 1
∑
converges by a limit comparison with the convergent
p-series
∞
1
∑ n2 .
n =1
∞
2n 2 − 1
∑ 3n5 + 2n + 1
n =1
converges by a limit comparison with the convergent
p-series
∞
1
∑ n3 .
n =1
© 2010 Brooks/Cole, Cengage Learning
302
Chapter 9
24. lim
n→∞
Infinite Series
n ⎡⎣( n + 1)2n −1 ⎤⎦
n
= lim
=1
n −1
n
→
∞
n +1
1 (2 )
28. lim
tan (1 n)
1n
n→∞
n
Therefore,
converges by a limit comparison with the convergent
geometric series
∑ tan⎜⎝ n ⎟⎠
n =1
∞
⎛1⎞
∞
diverges by a limit comparison with the divergent pseries
n −1
.
∞
( n k −1 ) ( n k
+ 1)
1n
n→∞
= lim
n→∞
nk
=1
n +1
n =1
∞ 13
29.
∞
30.
n =1
(
5 n +
n2 + 4
1n
n→∞
)=
lim
n→∞
n +
5
=
2
n2 + 4
31.
n2 + 4
Converges;
Direct comparison with convergent geometric series
∞
∞
⎛1⎞
∑ ⎜⎝ 5 ⎟⎠
1
∑ n.
n =1
∞
32.
∑ n3
n=3
1
−8
∞
Therefore,
Converges; limit comparison with
1
∑ n3
n=3
⎛1⎞
∑ sin⎜⎝ n ⎟⎠
∞
n =1
diverges by a limit comparison with the divergent p-series
33.
2n
∑ 3n − 2
n =1
∞
1
∑ n.
n =1
Diverges; nth-Term Test
lim
n→∞
∞
n
n =1
(−1 n2 ) cos(1 n) = lim cos⎛ 1 ⎞ = 1
sin(1 n)
27. lim
= lim
⎜ ⎟
n →∞
n →∞
n →∞
−1 n 2
1n
⎝n⎠
34.
1
7
1
+1
∑ 5n
n =1
diverges by a limit comparison with the divergent
harmonic series
∞
n
Geometric series with r = −
∞
5
n +
⎛ 1⎞
∑ 7⎜⎝ − 7 ⎟⎠
n=0
2
3
Converges;
5n
Therefore,
n =1
n =1
p-series with p =
1
∑ n.
∑
1
∑ n2 3
Diverges;
diverges by a limit comparison with the divergent p-series
∞
∞
n
=
n
∑
n =1
n k −1
∑ nk + 1
n =1
∞
26. lim
1
∑ n.
k
Therefore,
∞
⎛1⎞
n =1
n =1
25. lim
−1 n 2
⎛1⎞
= lim sec 2 ⎜ ⎟ = 1
n→∞
⎝n⎠
∑ (n + 1)2n −1
∑ ⎜⎝ 2 ⎟⎠
(−1 n2 ) sec2 (1 n)
n→∞
Therefore,
∞
= lim
⎛
1
∑ ⎜⎝ n + 1 −
n =1
1 ⎞
⎛ 1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞
⎟ = ⎜ − ⎟ +⎜ − ⎟ +⎜ − ⎟ +
n + 2⎠
⎝ 2 3⎠ ⎝ 3 4 ⎠ ⎝ 4 5⎠
=
2n
2
=
≠ 0
3n − 2
3
1
2
Converges; telescoping series
∞
35.
∑
n =1
n
(n
2
+ 1)
2
Converges; Integral Test
© 2010 Brooks/Cole, Cengage Learning
Section 9.4
36.
∞
3
∑ n(n + 3)
44. lim
n→∞
n =1
Converges; telescoping series
∞
⎛1
∑ ⎜⎝ n
−
n =1
n→∞
an
= lim nan . By given conditions lim nan is
n→∞
n→∞
1n
46.
∞
∑ an
n =1
∞
1
∑ n.
n =1
47.
38. If j < k − 1, then k − j > 1. The p-series with
p = k − j converges and because
P ( n) Q( n)
1n
k− j
= L > 0, the series
∞
P( n)
∞
48.
P( n)
1
2
3
4
5
+ +
+
+
+
2
5 10 17
26
=
1 1
1
1
1
40.
+ +
+
+
+
3 8 15
24 35
n =1
=
∞
1
∑ 200 + 10n
n=0
1
1
1
1
+
+
+
=
201 204
209
216
1
1
1
1
+
+
+
+
201 208
227
264
∞
n =1
n
∑ n2 + 1,
n =1
∞
1
∑ 200 + n2
n =1
=
∞
1
∑ 200 + n3
n =1
∞
1
n =1
51. See Theorem 9.13, page 628. One example is
∞
∑
∞
1
,
= ∑ 2
n=2 n − 1
n=2
∞
∑
n=2
52.
1
+1
1
1
1
converges because 2
< 2 and
n +1
n
+1
∑ n2 converges (p-series).
which converges because the degree of the numerator is
two less than the degree of the denominator.
∞
1
1
1
+
+
+
200
210
220
∑ n2
∞
which diverges because the degree of the numerator is
only one less than the degree of the denominator.
∑ n3
n =1
49. Some series diverge or converge very slowly. You
cannot decide convergence or divergence of a series by
comparing the first few terms.
diverges by the Limit Comparison Test.
41.
1
50. See Theorem 9.12, page 626. One example is
n =1
39.
∞
∑ 200n
converges
n =1
∑ Q( n)
=
converges
∑ Q( n)
converges by the Limit Comparison Test. Similarly, if
j ≥ k − 1, then k − j ≤ 1 which implies that
1
1
1
+
+
+
200
400
600
diverges
diverges by a limit comparison with the p-series
lim
1
diverges, (harmonic)
finite and nonzero. Therefore,
n→∞
∞
∑ ln n diverges.
n=2
45.
37. lim
303
n
1
= lim
= lim n = ∞ ≠ 0
n
→
∞
n→∞
ln n
1n
Therefore,
1 ⎞
⎟
n + 3⎠
Comparisons of Series
1
1 n −1
diverges because lim
= 1 and
n→∞
1 n
n −1
1
diverges (p-series).
n
1.0
Terms of
Σ an
0.8
converges because the degree of the numerator is three
less than the degree of the denominator.
n=1
0.6
Terms of
2
Σ an
0.4
n=1
0.2
∞
42.
n2
+1
n
∑ n3
n =1
4
diverges because the degree of the numerator is only one
less than the degree of the denominator.
8
12
16
20
For 0 < an < 1, 0 < an 2 < an < 1. So, the lower terms
are those of Σ an 2 .
⎛ n3 ⎞
n4
1
= lim
=
≠ 0
43. lim n⎜ 4
⎟
4
n→∞
n
→
∞
5n + 3
5
⎝ 5n + 3 ⎠
Therefore,
∞
n3
diverges.
+3
∑ 5n 4
n =1
© 2010 Brooks/Cole, Cengage Learning
304
Chapter 9
∞
1
∑
53. (a)
Infinite Series
( 2n
n =1
− 1)
∞
∑ 4n 2
=
2
n =1
1
− 4n + 1
converges because the degree of the numerator is two less than the degree of the denominator. (See Exercise 38.)
(b)
n
5
10
20
50
100
Sn
1.1839
1.2087
1.2212
1.2287
1.2312
∞
1
∑
(c)
( 2n
n=3
− 1)
∞
∑
(d)
n = 10
1
( 2n
− 1)
π2
=
2
π2
=
2
− S 2 ≈ 0.1226
8
− S9 ≈ 0.0277
8
54. This is not correct. The beginning terms do not affect the
convergence or divergence of a series. In fact,
1
1
+
+
1000 1001
∞
∞
1
1
∑ n diverges (harmonic)
n = 1000
1
= ∑ 2 converges ( p -series).
n
n =1
63.
n =1
1
⎛ 1 ⎞⎛ 1 ⎞
n =1
1
∑ n5 .
1
2
=
1
∑ n4 .
65. Suppose lim
n→∞
∞
∑ cn converges.
an ≤ bn + cn , but
=
∑ n2 converge, and therefore, so does
⎛1⎞
58. False. Let an = 1 n, bn = 1 n, cn = 1 n 2 . Then,
1
∑ n2 and ∑ n3 both converge, and therefore, so does
∑ ⎜⎝ n 2 ⎟⎠
an
= 0 and Σ bn converges.
bn
From the definition of limit of a sequence, there exists
M > 0 such that
an
−0 <1
bn
59. True
∞
∑ an could converge or diverge.
n =1
For example, let
∞
∑ bn
∞
∑
=
n =1
1
<
n
1
and
n
1
<
n2
∞
n =1
1
, which diverges.
n
1
∑ n diverges, but
n =1
1
and
n
∞
1
∑ n2 converges.
n =1
∞
61. Because
∞
∑ an 2 converges by Exercise 61.
∑ ⎜⎝ n 2 ⎟⎜
3⎟
⎠⎝ n ⎠
64.
n =1
60. False.
=
n =1
1
57. True
0 <
∑ an an
1
1
and bn = 2 . 0 < an ≤ bn and both
n3
n
56. True
0 <
∞
∞
∑ n3 and ∑ n2 converge.
n =1
∞
∑ an converges, then
n =1
∞
=
1
1
and 1 + + +
4 9
55. False. Let an =
62. Because
∑ bn converges,
n =1
lim bn = 0. There exists N
n→∞
whenever n > M . So, an < bn for n > M . From the
Comparison Test, Σ an converges.
66. Suppose lim
n→∞
an
= ∞ and Σ bn diverges. From the
bn
definition of limit of a sequence, there exists
M > 0 such that
an
>1
bn
for n > M . So, an > bn for n > M . By the
Comparison Test, Σ an diverges.
such that bn < 1 for n > N . So, anbn < an for
n > N and
∞
∑ anbn converges by comparison to the
n =1
convergent series
∞
∑ an .
i =1
© 2010 Brooks/Cole, Cengage Learning
Section 9.4
∑ an
67. (a) Let
=
∑
1
(n
+ 1)
3
, and
∑ bn
=
1
∑ n2 ,
converges.
3
1 ⎡( n + 1) ⎤
a
n2
⎦ = lim
= 0
lim n = lim ⎣
2
n→∞ b
n→∞
n→∞ n + 1 3
1 (n )
n
(
)
∞
∑
By Exercise 65,
n =1
∑ an
(b) Let
=
1
(n
+ 1)
3
1
, and
nπ n
∑
converges.
∑ bn
=
1
converges.
lim
n→∞
(
)
1
nπ n
an
1
= lim
= lim
= 0
n→∞
n→∞
bn
n
1 (π n )
∞
1
converges.
nπ n
∑
n =1
ln n
68. (a) Let ∑ an = ∑
, and
n
lim
1
∑ bn = ∑ n , diverges.
(ln n) n = lim ln n = ∞
an
= lim
→
∞
n
n→∞
1n
bn
By Exercise 66,
∞
ln n
diverges.
n =1 n
∑
1
, and
(b) Let ∑ an = ∑
ln n
lim
n→∞
x ≈ 5503.66. That is,
ln n < n1 4 for n > 5504
which implies that
ln n
1
< 5 4 for n > 5504.
n3 2
n
1
∑ bn = ∑ n , diverges.
an
n
= lim
= ∞
→
∞
n
bn
ln n
By Exercise 66,
ln n
∑ n3 2
converges by direct comparison.
72. The series diverges. For n > 1,
n < 2n
1n
n
1
∑ ln n diverges.
n→∞
lim
an
< 2
1
1
>
n1 n
2
1
1
>
n + 1) n
(
2
n
n
Because
1
If an ≥
1 ( n + 1)
1
2
n +1
⎛ 1 ⎞
, then an1 (n + 1) ≥ ⎜ n + 1 ⎟
⎝2 ⎠
an
≤ 2an .
1 ( n + 1)
=
1
, and
2
=
1
, and
2n
an
⎛ 1 ⎞
, then an n (n + 1) ≤ ⎜ n + 1 ⎟
2n +1
⎝2 ⎠
1
combining, an n (n + 1) ≤ 2an + n .
2
If an ≤
Because
= 1 and ∑ an
1
∑ 2n diverges, so does ∑ n(n +1) n .
73. Consider two cases:
for sufficiently large n. Because
n→∞
1
n =1
∞
an n (n + 1) =
69. Because lim an = 0, the terms of Σ sin ( an ) are positive
sin ( an )
∞
∑ n5 4 is a convergent p-series,
n =1
By Exercise 65,
n→∞
305
71. First note that f ( x) = ln x − x1 4 = 0 when
Because
∑πn,
Comparisons of Series
1
∞
⎛
∑ ⎜⎝ 2an
n =1
+
n ( n + 1)
1⎞
⎟ converges, so does
2n ⎠
∞
∑ an n (n +1)
n =1
by the Comparison Test.
converges, so does Σ sin ( an ).
70.
∞
1
∑1 + 2 +
n =1
+ n
=
∞
n =1
=
1
∑ ⎡n(n + 1)⎤
∞
⎦ 2
⎣
2
∑ n(n + 1)
n =1
Because Σ1 n 2 converges, and
lim
n→∞
2 ⎡⎣n( n + 1)⎤⎦
2n 2
= lim
= 2,
2
n → ∞ n( n + 1)
1 (n )
1
∑1 + 2 +
+ n
converges.
© 2010 Brooks/Cole, Cengage Learning
306
Chapter 9
Infinite Series
Section 9.5 Alternating Series
∞
1.
6
∑ n2
4.
n =1
S1 = 6
Matches (b).
(−1)n −1 6
n
n =1
5.
2
∞
10
∑ n2n
n =1
S1 = 6
S1 = 5
S 2 = 4.4
S 2 = 6.25
S3 ≈ 5.1667
Matches (e).
Matches (f).
3.
n!
S3 = 2.0
Matches (d).
∞
n =1
S 2 = 1.5
S3 ≈ 8.1667
∑
(−1)n −1 3
S1 = 3
S 2 = 7.5
2.
∞
∑
∞
6.
3
∑ n!
∞
∑
(−1) 10
n =1
n
n2n
n =1
S1 = 5
S1 = 3
S 2 = 3.75
S 2 = 4.5
Matches (c).
S3 = 5.0
Matches (a).
7.
∞
(−1)
n −1
∑ 2n − 1
=
n =1
(a)
(b)
π
4
≈ 0.7854
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.6667
0.8667
0.7238
0.8349
0.7440
0.8209
0.7543
0.8131
0.7605
1.1
11
0
0.6
(c) The points alternate sides of the horizontal line y =
π
4
that represents the sum of the series.
The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term of the series.
© 2010 Brooks/Cole, Cengage Learning
Section 9.5
8.
∞
(−1)
n −1
∑ (n − 1)!
=
n =1
(a)
(b)
Alternating Series
307
1
≈ 0.3679
e
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0
0.5
0.3333
0.375
0.3667
0.3681
0.3679
0.3679
0.3679
2
0
11
0
1
that represents the sum of the series.
e
(c) The points alternate sides of the horizontal line y =
The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
∞
9.
∑
(−1)
n −1
n2
n =1
(a)
(b)
π2
=
≈ 0.8225
12
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.75
0.8611
0.7986
0.8386
0.8108
0.8312
0.8156
0.8280
0.8180
1.1
0
11
0.6
(c) The points alternate sides of the horizontal line y =
π2
12
that represents the sum of the series.
The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next term in the series.
10.
∞
(−1)n −1
∑ (2n − 1)!
= sin (1) ≈ 0.8415
n =1
(a)
(b)
n
1
2
3
4
5
6
7
8
9
10
Sn
1
0.8333
0.8417
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
0.8415
2
0
11
0
(c) The points alternate sides of the horizontal line y = sin (1) that represents the sum of the series.
The distance between successive points and the line decreases.
(d) The distance in part (c) is always less than the magnitude of the next series.
© 2010 Brooks/Cole, Cengage Learning
308
Chapter 9
∞
11.
∑
Infinite Series
(−1)n +1
∞
17.
n +1
n =1
n =1
1
1
<
= an
n + 2
n +1
1
lim an = lim
= 0
n→∞
n→∞ n + 1
an + 1 =
∞
∑
(−1)
n +1
n
lim
n→∞
lim
n→∞
n
1
=
3n + 2
3
( −1)n
∑ ln(n + 1)
n =1
∞
lim
∞
∑
(−1)
n→∞
n
1
1
< n = an
3n + 1
3
an + 1 =
lim
n→0
∞
19.
lim
n→∞
Converges by Theorem 9.14
∞
⎛ −1 ⎞
∑ ⎜⎝ 3 ⎟⎠
14.
∑
(−1)
is a convergent geometric series)
∞
20.
an + 1 =
1
1
< n = an
en +1
e
(−1) (5n
lim
n→∞
n
21.
∞
5n − 1
5
=
4n + 1
4
n
n
n =1
lim
n→∞
n
= ∞
ln ( n + 1)
Diverges by nth-Term test
n
= 0
n2 + 5
(−1)n
n
an + 1 =
4n + 1
( −1)
∞
∑
n =1
− 1)
∑ ln(n + 1)
−( x 2 − 5)
x
′
,
=
< 0 for x ≥ 3
f
x
(
)
2
x2 + 5
( x2 + 5)
Converges by Theorem 9.14
lim
n →∞
Diverges by nth-Term test
16.
n2 + 5
n →∞
n
⎛ −1 ⎞
(Note: ∑ ⎜ ⎟ is a convergent geometric series)
n =1 ⎝ e ⎠
∑
n2
=1
n +5
2
(−1)n +1 n
lim
∞
n =1
+5
So, an + 1 < an for n ≥ 3
Converges by Theorem 9.14
15.
n2
Let f ( x) =
1
= 0
lim
n → ∞ en
∞
∑
n =1
n
en
n =1
n
Diverges by nth-Term test
n
n =1
∞
(−1)
∑ n2
n =1
1
= 0
3n
(Note:
1
= 0
ln ( n + 1)
Converges by Theorem 9.14
3n
n =1
1
1
<
= an
ln ( n + 2)
ln ( n + 1)
an + 1 =
Diverges by nth-Term test
13.
1
= 0
3n + 2
Converges by Theorem 9.14
18.
3n + 2
n =1
1
1
<
= an
3( n + 1) + 2
3n + 2
an + 1 =
Converges by Theorem 9.14
12.
( −1)n +1
∑ 3n + 2
1
<
n +1
1
= an
n
1
= 0
n
Converges by Theorem 9.14
22.
∞
∑
(−1)n +1 n 2
n =1
lim
n →∞
n2 + 4
n2
=1
n + 4
2
Diverges by nth-Term test
© 2010 Brooks/Cole, Cengage Learning
Section 9.5
23.
(−1)n +1(n + 1)
ln ( n + 1)
n =1
∞
∑
29.
n +1
1
= lim
= lim ( n + 1) = ∞
lim
n → ∞ ln ( n + 1)
n → ∞ 1 ( n + 1)
n→∞
∞
(−1)n
n=0
n!
∑
lim
n→∞
∞
24.
∑
(−1)
n +1
ln ( n + 1)
an + 1
lim
ln ⎡( n + 1) + 1⎤⎦
ln ( n + 1)
= ⎣
<
for n ≥ 2
n +1
(n + 1) + 1
ln ( n + 1)
= lim
n +1
n→∞
1 ( n + 1)
⎡ ( 2n − 1)π ⎤
⎥ =
2
⎣
⎦
∞
∞
∑ sin ⎢
n =1
∑ (−1)
an + 1 =
lim
n→∞
n =1
∞
31.
∑
∑ cos nπ =
n =1
lim
n→∞
1
∞
32.
n
∑ (−1)
n =1
an + 1 =
lim
n→∞
∑
lim
n→∞
∞
(−1)n
n =1
n
3
n
n
12
n =1
∑
n +1
n
<
for n ≥ 2
+ 1) + 2
n + 2
n
= 0
n + 2
(−1)n +1
n =1
∞
=
(n
Converges by Theorem 9.14
Diverges by the nth-Term Test
∑ n cos nπ
n
n + 2
an + 1 =
Converges by Theorem 9.14
∞
n +1
n +1
∞
1
1
<
= an
+ 3)!
(2n + 1)!
1
= 0
2
n
( + 1)!
(−1)
n =1
∞ −1
⎡ ( 2n − 1)π ⎤
( )
1
26. ∑ sin ⎢
⎥ = ∑
n
n
2
n =1
n =1
⎣
⎦
1
1
<
= an
an + 1 =
n +1
n
1
= 0
lim
n→∞ n
∞
28.
( 2n
Converges by Theorem 9.14
n +1
Diverges by the nth-Term Test
27.
n
∑ (2n + 1)!
n=0
= 0
1
n→∞
(−1)
∞
30.
Converges by Theorem 9.14
25.
1
= 0
n!
Converges by Theorem 9.14
n +1
n =1
309
1
1
<
= an
n
n
+
1
!
!
(
)
an + 1 =
Diverges by the nth-Term Test
Alternating Series
n
= lim n1 6 = ∞
n→∞
n1 3
Diverges by the nth-Term Test
1
1
<
= an
n +1
n
1
= 0
n
Converges by Theorem 9.14
33.
(−1)n +1 n!
∑ 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n − 1)
n =1
(n + 1)!
an + 1 =
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1)( 2n
∞
lim an = lim
n →∞
n →∞
+ 1)
=
n!
n +1
⎛ n +1⎞
⋅
= an ⎜
⎟ < an
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1) 2n + 1
⎝ 2n + 1 ⎠
n!
1⋅ 2⋅3⋅⋅⋅ n
n ⎤
1
⎡3 4 5
= lim
= lim 2 ⎢ ⋅ ⋅ ⋅ ⋅ ⋅
⋅
= 0
n →∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2 n − 1)
n →∞ ⎣ 3
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 1)
5 7
2n − 3 ⎥⎦ 2n − 1
Converges by Theorem 9.14
© 2010 Brooks/Cole, Cengage Learning
310
Chapter 9
∞
34.
∑ (−1)
n +1 1 ⋅
3⋅5
1⋅ 4 ⋅ 7
n =1
an +1 =
Infinite Series
1⋅ 3⋅ 5
1⋅ 4 ⋅ 7
(2n − 1)
(3n − 2)
(2n − 1)(2n + 1)
(3n − 2)(3n + 1)
⎛ 2n + 1 ⎞
= an ⎜
⎟ < an
⎝ 3n + 1 ⎠
Converges by Theorem 9.14
35.
∞
∑
(−1) (2)
n =1
n +1
e −e
n
Let f ( x) =
f ′( x) =
(−1)
∞
∑
=
−n
e
n =1
n +1
2n
−2e (e
(e
+ 1)
2x
− 1)
2
n→∞
( 2e n )
36.
∑ en
n =1
f ′( x) =
∑
=
+ e− n
Let f ( x) =
∞
∞
41.
∑
n=0
< 0.
n +1
(−1)n
n=0
n!
=1−1+
1 1
1
1
1
− +
−
+
2 6
24 120
720
(7 terms. Note that the sum begins with n = 0. )
2e x
. Then
e +1
2x
(e
+ 1)
2x
2
< 0 for x > 0.
∞
42.
∑
n=0
2( −1)
4( −1)
4
∑
(−1)n
n
2 n!
=1−
43.
∞
(−1)n
∑ (2n + 1)!
n=0
(a) By Theorem 9.15,
≈ 2.7067
4
≤ a7 =
≈ 1.9236
ln 8
0.7831 ≤ S ≤ 4.6303
1 1
1
1
+ −
+
≈ 0.607.
2 8 48 348
(5 terms. Note that the sum begins with n = 0. )
RN ≤ aN + 1 =
n =1
1
< 0.001.
( N + 1)!
(b) Approximate the series by
n=0
n +1
2
N +1
This inequality is valid when N = 4.
≈ 0.7333
∑ ln(n + 1)
R6 = S − S6
2n n!
Rn ≤ aN + 1 =
n
n!
(−1)n
(a) By Theorem 9.15,
2
= 0.002778
R6 = S − S6 ≤ a7 =
6!
0.7305 ≤ S ≤ 0.7361
38. S6 =
1
< 0.001.
+ 1)!
≈ 0.368.
0.7333 − 0.002778 ≤ S ≤ 0.7333 + 0.002778
6
6
∑
( 2e )
n
The series converges by Theorem 9.14.
∑
(N
(b) Approximate the series by
2n
n=0
n!
e2 n + 1
2e2 x (1 − e 2 x )
5
n
This inequality is valid when N = 6.
2e n
2e n
1
= lim
= lim n = 0
n
n→∞ e
→
∞
e +1
2e 2 n
37. S6 =
(−1)
RN ≤ aN + 1 =
an + 1 < an .
lim
7
≈ 0.05469
27
(a) By Theorem 9.15,
(−1)
n =1
= 0.1875
0.1328 ≤ S ≤ 0.2422
So, f ( x) is decreasing for x > 0 which implies
n→∞
2n
R6 = S − S6 ≤ a7 =
−1
2e n
2e n
1
= lim 2 n = lim n = 0.
n → ∞ 2e
n→∞ e
e −1
2( −1)
∑
(−1)n +1 n
n =1
The series converges by Theorem 9.14.
∞
6
40. S6 =
2n
n +1
3
≈ 0.0612
49
2.3713 ≤ S ≤ 2.4937
So, f ( x) is decreasing. Therefore, an + 1 < an , and
lim
= 2.4325
n2
R6 = S − S6 ≤ a7 =
2e x
. Then
e −1
2x
n +1
2.4325 − 0.0612 ≤ S ≤ 2.4325 + 0.0612
2x
x
3( −1)
n =1
2n − 1 ⎤ 1
=0
3n − 5 ⎥⎦ 3n − 2
⎡5 7 9
lim an = lim 3⎢ ⋅ ⋅
n →∞
n →∞ ⎣ 4 7 10
6
∑
39. S6 =
1
< 0.001.
⎣⎡2( N + 1) + 1⎤⎦!
This inequality is valid when N = 2.
(b) Approximate the series by
2
(−1)n
∑ (2n + 1)!
n=0
=1−
1
1
+
≈ 0.842.
6 120
(3 terms. Note that the sum begins with n = 0. )
© 2010 Brooks/Cole, Cengage Learning
Section 9.5
44.
∞
(−1)
n
∞
∑ (2n)!
48.
n=0
∑
1
< 0.001.
(2 N + 2)!
RN ≤ aN +1 =
(−1)
=1−
n=0
Use 31 terms.
1
1
1
+
−
≈ 0.540.
2
24
720
49.
45.
∑
(−1)
RN ≤ aN + 1 =
∞
1
< 0.001.
N +1
50.
1000
(−1)
n =1
n
n +1
=1−
1
1 1
+ − +
2
3 4
−
1
1000
∞
∞
1
= N +1
< 0.001.
4 ( N + 1)
(b) Approximate the series by
n
4 n
=
47.
∑
n =1
( −1)
1
∞
52.
∑
(−1)
< 0.01
∞
∑
1
∑ n2
∞
53.
By Theorem 9.15,
∑
n =1
1
3
< 0.001
⇒ ( N + 1) > 1000 ⇒ N + 1 > 10.
converges absolutely.
is a convergent p-series.
n =1
∞
∑
n =1
n3
2n
n2
n =1
∞
(−1)n
n +1
Therefore,
3
5
is a convergent geometric series.
n =1
1
1
1
−
+
≈ 0.224.
4 32 192
( N + 1)
+ 1)
2n
n +1
RN ≤ aN + 1 =
(N
n
Therefore,
(3 terms)
∞
1
n =1
This inequality is valid when N = 3.
( −1)n +1
(−1)
∑ 2n
(a) By Theorem 9.15,
n =1
∑
n =1
4n n
3
n5
Use 2 terms.
(−1)n +1
∑
< 0.001.
This inequality is valid when N = 2.
51.
RN ≤ aN + 1
3
(−1)n +1
RN ≤ aN + 1 =
(1000 terms)
∑
1
2( N + 1) − 1
By Theorem 9.15,
≈ 0.693.
n =1
∑
n =1
(b) Approximate the series by
46.
−1
This inequality is valid when N = 7. Use 7 terms.
This inequality is valid when N = 1000.
∞
n +1
RN ≤ aN + 1 =
n +1
(a) By Theorem 9.15,
∑
(−1)
By Theorem 9.15,
n!
n =1
∞
∑ 2n3
n =1
(4 terms. Note that the sum begins with n = 0. )
∞
< 0.001
2
2
(b) Approximate the series by
∑ (2n)!
1
( N + 1)
⇒ ( N + 1) > 1000 ⇒ N = 31.
This inequality is valid when N = 3.
3
n2
By Theorem 9.15,
(a) By Theorem 9.15,
n
311
( −1)n +1
n =1
RN ≤ aN + 1 =
Alternating Series
(−1)
(−1)n +1 converges absolutely.
n2
n
n!
1
1
< 2 for n ≥ 4
n!
n
∞
and
1
∑ n2 is a convergent p-series.
n =1
Use 10 terms.
So,
∞
1
∑ n! converges, and
n =1
∞
(−1)n
n =1
n!
∑
converges absolutely.
© 2010 Brooks/Cole, Cengage Learning
312
Chapter 9
∞
54.
∑
(−1)n +1
∞
n =1
∞
∑
n
n3
∞
1
(−1)n +1
n
60.
(−1)
2
n = 1 ( n + 3)
∑
∞
∑
n =1
1
(n
+ 3)
2
∞
converges
61.
∑
2n + 3
= 2
n + 10
( −1)n
∑ n ln n
n=2
The series converges by the Alternating Series Test.
1
.
Let f ( x) =
x ln x
absolutely.
n =1
+ 3)
Therefore, the series diverges by the nth-Term Test.
converges by a limit comparison to the
(−1)n +1
∑ n+32
n =1 (
)
=1
2
n + 10
lim
∞
+ 1)
( −1)n +1 (2n
n→∞
1
p-series ∑ 2 . Therefore,
n =1 n
∞
∞
∑
n =1
∞
56.
(n
Therefore, the series diverges by the nth-Term Test.
n +1
∞
n2
n→∞
converges absolutely.
n3
∑
lim
n =1
n =1
55.
59.
∑ n5 2 is a convergent p-series. Therefore,
=
(−1)n +1 n2
2
n = 1 ( n + 1)
∞
n
n3
n =1
∑
Infinite Series
(−1)n +1
n +3
∞
∫2
The series converges by the Alternating Series Test. But,
the series
∞
1
dx = ⎡⎣ln (ln x)⎤⎦ 2 = ∞
x ln x
∞
1
∑n + 3
n =1
1
∑ n ln n diverges.
By the Integral Test,
∞
n=2
∞
diverges by comparison to
∞
So, the series
1
∑ n.
(−1)
n
∑ n ln n
converges conditionally.
n=2
n =1
∞
∑
Therefore,
(−1)n +1 converges conditionally.
n =1
∞
57.
∑
(−1)
∞
62.
n +3
∞
1
n
n =1
n =1
n n
∞
n =1
1
n
en
converges by a comparison to the convergent
2
∞
⎛1⎞
n
∑ ⎜⎝ e ⎟⎠ . Therefore, the given series
n=0
∞
(−1)n +1
∑n
1
converges absolutely.
63.
∞
∑
∞
2
geometric series
( −1)
∑ n3
n=2
is a divergent p-series. Therefore, the series converges
conditionally.
58.
en
n=0
The given series converges by the Alternating Series
Test, but does not converge absolutely because
∑
(−1)n
n=0
∑
n +1
n
n =1
∑
∞
∑ n3
n=2
n
n
−5
n
converges by a limit comparison to the
−5
∞
p-series
1
∑ n 2 . Therefore, the given series converges
n=2
=
∞
1
absolutely.
∑ n3 2 which is a convergent p-series.
n =1
Therefore, the given series converges absolutely.
∞
64.
∑
n =1
∞
( −1)
n +1
n4 3
1
∑ n4 3 is a convergent p-series. Therefore, the given
n =1
series converges absolutely.
© 2010 Brooks/Cole, Cengage Learning
Section 9.5
65.
( −1)n
∞
69.
n=0
∞
∞
1
313
(−1)n
n2
n =1
is convergent by comparison to the convergent geometric
series
Therefore, the given series converges absolutely.
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠
n=0
is a convergent p-series.
n =1
n
sin ⎡⎣( 2n − 1)π 2⎤⎦
=
n
n =1
∞
70.
∑
∞
( −1)n +1
n =1
n
∑
because
The given series converges by the Alternating Series
Test, but
1
1
< n for n > 0.
(2n + 1)! 2
∑
Therefore, the given series converges absolutely.
(−1)n
∞
∑
n + 4
n=0
The given series converges by the Alternating Series
Test, but
∞
1
n + 4
∑
n=0
∞
n =1
cos nπ
∑ n+1 =
n=0
∞
(−1)
∞
n
∑n+1
n=0
The given series converges by the Alternating Series
Test, but
∑
cos nπ
n=0
∞
1
= ∑
n +1
n
+1
n=0
diverges by a limit comparison to the divergent harmonic
series,
∞
1
∑ n.
n =1
lim
cos nπ
(n
+ 1)
1n
= 1, therefore, the series
converges conditionally.
∞
∑ (−1)
n +1
71. An alternating series is a series whose terms alternate in
sign.
72. See Theorem 9.14.
(Theorem 9.15)
74. ∑ an is absolutely convergent if ∑ an converges.
75. (b). The partial sums alternate above and below the
horizontal line representing the sum.
76. (a) False. For example, let an =
∑ an
Then
arctan n
lim arctan n =
π
2
∑
=
and
∑ ( − an )
But,
∑
( −1)
n
n
converges
n
∑
=
(−1)n .
(−1)n +1 converges.
n
1
∑ n diverges.
an =
∑
an converged, then so would
∑ an by Theorem 9.16.
77. True. S100 = −1 +
1
2
−
1
3
+
+
1
100
1 is negative,
Because the next term − 101
S100 is an
overestimate of the sum.
n =1
n→∞
1
n =1
is a divergent p-series. Therefore, the series converges
conditionally.
(b) True. For if
n→∞
∞
∑n
∑ an converges.
Therefore, the given series converges conditionally.
∞
n =1
sin ⎡⎣( 2n − 1)π 2⎤⎦
=
n
∑ an is conditionally convergent if ∑ an diverges, but
1
.
n
∑
∞
73. S − S N = RN ≤ aN +1
diverges by a limit comparison to the divergent p-series
68.
1
∞
∑
∑ n2
∞
67.
∑
∑ (2n + 1)!
n=0
66.
cos nπ
=
n2
n =1
∞
∑ (2n + 1)!
Alternating Series
≠ 0
Therefore, the series diverges by the nth-Term Test.
78. False. Let
∑ an
=
∑ bn
=
∑
(−1)n .
n
Then both converge by the Alternating Series Test. But,
1
∑ anbn = ∑ n , which diverges.
© 2010 Brooks/Cole, Cengage Learning
314
79.
Chapter 9
∞
∑ (−1)
n
n =1
Infinite Series
1
np
84. (a)
If p = 0, then
∞
∑ (−1)
n
If p < 0, then
∑ (−1)
n
n
−p
xn
n =1 n
converges absolutely (by comparison) for
−1 < x < 1, because
diverges.
n =1
∞
∞
∑
xn
< x n and
n
diverges.
n =1
If p > 0, then lim
n→∞
1
an + 1 =
(n
+ 1)
p
<
is a convergent geometric series for −1 < x < 1.
1
= 0 and
np
(b) When x = −1, you have the convergent alternating
series
1
= an .
np
∞
∑
Therefore, the series converges for p > 0.
∞
∑ xn
( −1)n .
n =1
n
When x = 1, you have the divergent harmonic
series 1 n. Therefore,
1
80. ∑ ( −1)
n + p
n =1
n
∞
xn
converges conditionally for x = −1.
n =1 n
∑
Assume that n + p ≠ 0 so that an = 1 ( n + p ) are
defined for all n. For all p,
lim an = lim
n→∞
n→∞
1
= 0
n + p
1
1
<
< an .
n +1+ p
n + p
an + 1 =
85. (a) No, the series does not satisfy an + 1 ≤ an for all n.
1
1
< .
For example,
9
8
(b) Yes, the series converges.
1 1
1
− +
+ n
2 3
2
1⎞
⎛1
= ⎜ +
+ n⎟ −
2 ⎠
⎝2
S2n =
Therefore, the series converges for all p.
81. Because
∞
∑ an
1
⎛
= ⎜1 + +
2
⎝
n =1
converges you have lim an = 0. So, there must exist
+
−
1
3n
⎛1
⎜ +
⎝3
+
1⎞
⎟
3n ⎠
1⎞ ⎛
1
⎟ − ⎜1 + +
2n ⎠ ⎝
3
+
1⎞
⎟
3n ⎠
As n → ∞,
n→∞
an N > 0 such that aN < 1 for all n > N and it
S2n →
follows that an 2 ≤ an for all n > N . So, by the
1
1
3
1
−
= 2−
= .
1 − (1 2) 1 − (1 3)
2
2
Comparison Test,
86. (a) No, the series does not satisfy an + 1 ≤ an :
∞
∑ an 2
∞
∑ (−1)
n =1
converges. Let an = 1 n to see that the converse is false.
82.
∞
∑
( −1)
n
n =1
83.
∞
1
∑ n2
n =1
n −1
converges, but
∞
1
<
8
1
∑ n diverges.
n =1
converges, and so does
∞
an = 1 −
1
1
1
+
−
+
8
64
3
and
1
.
3
(b) No, the series diverges because
∑
1
diverges.
n
1
∑ n4 .
n =1
n +1
n =1
87.
∞
10
∑ n3 2
n =1
∞
= 10∑
n =1
1
,
n3 2
convergent p-series
88.
∞
∑ n2
n =1
3
+5
converges by limit comparison to convergent p-series
1
∑ n2 .
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
89. Diverges by nth-Term Test
The Ratio and Root Tests
315
94. Converges (conditionally) by Alternating Series Test
lim an = ∞
95. Converges (absolutely) by Alternating Series Test
n→∞
90. Converges by limit comparison to convergent geometric
1
series ∑ n .
2
96. Diverges by comparison to Divergent Harmonic Series:
91. Convergent geometric series
97. The first term of the series is zero, not one. You cannot
regroup series terms arbitrarily.
(r
=
7
8
ln n
1
> for n ≥ 3
n
n
)
<1
98. Rearranging the terms of an alternating series can change
the sum.
92. Diverges by nth-Term Test
3
2
lim an =
n→∞
(
93. Convergent geometric series r = 1
)
e or Integral
Test
1
1 1
+ − +
2
3 4
1 1
1
1
1
1
1
1
S =1+ − + + − + +
− +
3 2
5
7
4
9 11 6
99. s = 1 −
1
1 1
1 1
+ − + − +
2
3 4
5 6
1
1
1 1
1
=
− + − +
−
2
4
6 8 10
(i) s4 n = 1 −
1
s2 n
2
Adding: s4 n +
(ii) lim sn = s
n→∞
1
1 1 1
1
1
s2 n = 1 + − + + − +
2
3 2 5 7
4
+
1
1
1
+
−
= s3n
4n − 3 4 n − 1 2n
(In fact, s = ln 2. )
s ≠ 0 because s >
1
.
2
S = lim S3n = s4 n +
n→∞
1
1
−
4n − 1 4n
1
1
+
−
4n − 2
4n
+
1
1
3
s2 n = s + s = s
2
2
2
So, S ≠ s.
Section 9.6 The Ratio and Root Tests
1.
(n + 1)!
(n + 2)!
2.
(2k − 2)!
(2k )!
=
=
(n
+ 1)( n)( n − 1)( n − 2)!
= ( n + 1)( n)( n − 1)
( n − 2)!
1
(2k − 2)!
=
2
k
2
k
−
1
2
k
−
2
!
2
k
2
( )(
)(
) ( )( k − 1)
© 2010 Brooks/Cole, Cengage Learning
316
Chapter 9
Infinite Series
3. Use the Principle of Mathematical Induction. When k = 1, the formula is valid because 1 =
( 2n
1⋅3⋅5
− 1) =
(2(1))!. Assume that
21 ⋅ 1!
( 2n)!
2n n!
and show that
( 2n
1⋅3⋅5
− 1)( 2n + 1) =
( 2n + 2)! .
2n + 1 ( n + 1)!
To do this, note that:
( 2n
1⋅3⋅5
− 1)( 2n + 1) = ⎡⎣1 ⋅ 3 ⋅ 5
( 2n
− 1)⎤⎦ ( 2n + 1)
(2n)! ⋅
(2n + 1) (Induction hypothesis)
2n n!
(2n)!(2n + 1) ⋅ (2n + 2)
=
2n n!
2( n + 1)
=
=
=
( 2n)!(2n
+ 1)( 2n + 2)
2n + 1 n!( n + 1)
( 2n + 2)!
2n + 1 ( n + 1)!
The formula is valid for all n ≥ 1.
4. Use the Principle of Mathematical Induction. When k = 3, the formula is valid because
1
1⋅3⋅5
( 2n
− 5)
=
233!(3)(5)
1
=
= 1. Assume that
1
6!
2n n!( 2n − 3)( 2n − 1)
(2n)!
and show that
1⋅3⋅5
2n + 1 ( n + 1)!( 2n − 1)( 2n + 1)
1
=
.
(2n − 5)(2n − 3)
(2n + 2)!
To do this, note that:
1
1⋅3⋅5
( 2n
− 5)( 2n − 3)
=
=
=
=
=
1
⋅
1
(2n − 5) (2n − 3)
n
2 n! ( 2n − 3) ( 2n − 1)
1
⋅
(2n)!
(2n − 3)
1⋅3⋅5
2n n!( 2n − 1)
(2n + 1)(2n + 2)
⋅
(2n)!
(2n + 1)(2n + 2)
n
2 ( 2)( n + 1)n!( 2n − 1)( 2n + 1)
(2n)!(2n + 1)(2n + 2)
n +1
2 ( n + 1)!( 2n − 1)( 2n + 1)
(2n + 2)!
The formula is valid for all n ≥ 3.
∞
5.
∑ n( 34 )
n
( 34 ) + 2(169 ) +
=1
n =1
S1 =
3
,
4
S2 ≈ 1.875
Matches (d).
∞
6.
n
3
9 ⎛1⎞
⎛ 3⎞ ⎛ 1 ⎞
∑ ⎜ ⎟ ⎜ ⎟ = 4 + 16 ⎜⎝ 2 ⎟⎠ +
n = 1 ⎝ 4 ⎠ ⎝ n! ⎠
S1 =
3
, S2 ≈ 1.03
4
Matches (c).
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
7.
∞
(−3)n +1
n =1
n!
Matches (f).
Matches (a).
(−1) 4
∑ (2n)!
n =1
n −1
33
+
2
9.
∞
4
4
=
−
+
2
24
10.
∑ 4e− n
=
= 4+
4
+
e
S1 = 4
Matches (e).
Matches (b).
(n + 1) (5 8)
an + 1
= lim
n
n→∞
an
n 2 (5 8)
2
11. (a) Ratio Test: lim
n→∞
(c)
⎞
n=0
S1 = 2
(b)
4n
∑ ⎜⎝ 5n − 3 ⎟⎠
317
2
S1 = 9
= 9−
⎛
n
4 ⎛8⎞
+⎜ ⎟ +
2 ⎝7⎠
n =1
S1 = 2, S 2 = 3.31
∑
∞
8.
∞
The Ratio and Root Tests
n +1
2
5
⎛ n + 1⎞ 5
= lim ⎜
=
< 1. Converges
⎟
n→∞ ⎝
n ⎠ 8
8
n
5
10
15
20
25
Sn
9.2104
16.7598
18.8016
19.1878
19.2491
20
0
12
0
(d) The sum is approximately 19.26.
(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum
of the series.
(n
12. (a) Ratio Test: lim
n→∞
(b)
(c)
an + 1
= lim
n→∞
an
+ 1) + 1
(n + 1)! = lim ⎛ n2 + 2n + 2 ⎞⎛ 1 ⎞ = 0 < 1. Converges
⎜
⎟⎜
⎟
2
n→∞
n2 + 1
⎝ n + 1 ⎠⎝ n + 1 ⎠
n!
2
n
5
10
15
20
25
Sn
7.0917
7.1548
7.1548
7.1548
7.1548
10
0
11
0
(d) The sum is approximately 7.15485
(e) The more rapidly the terms of the series approach 0, the more rapidly the sequence of the partial sums approaches the sum
of the series.
∞
13.
1
∑ 2n
n =1
n +1
lim
n →∞
an +1
12
= lim
n →∞ 1 2n
an
= lim
n →∞
2n
1
=
<1
2 n +1
2
Therefore, the series converges by the Ratio Test.
© 2010 Brooks/Cole, Cengage Learning
318
Chapter 9
∞
14.
Infinite Series
∞
1
∑ n!
20.
n =1
n =1
1 ( n + 1)!
an +1
= lim
n →∞
an
1 n!
lim
n →∞
n3
∑ 4n
n→∞
n!
1
= lim
= lim
= 0
n →∞ ( n + 1)!
n →∞ n + 1
= lim
∞
21.
lim
n→∞
n→∞
an +1
3
n!
3
= lim
⋅
= lim
= 0
n →∞ ( n + 1)! 3n
n →∞ n + 1
an
⎛6⎞
∑ n⎜⎝ 5 ⎟⎠
lim
= lim
n→∞
n→∞
n +1
⎛ 10 ⎞
∑ n⎜⎝ 9 ⎟⎠
23.
n +1
n=0
n!
∑
n→∞
n→∞
∞
∑
(−1)n −1 (3 2)n
n =1
n +1
= lim
n →∞
n +1
=14 <1
4n
Therefore, the series converges by the Ratio Test.
2
= 0
n +1
Therefore, by the Ratio Test, the series converges.
24.
n =1
an + 1
2n + 1
n!
= lim
⋅ n
n → ∞ ( n + 1)!
an
2
= lim
n
∑ 4n
(n + 1) 4
an +1
= lim
n
n →∞
an
n4
n +3
n + 2
≤
= an
+ 1)( n + 2)
n( n + 1)
n + 2
= 0
n( n + 1)
(−1)n 2n
lim
Therefore, the series diverges by the Ratio Test.
lim
(n
∞
n + 1⎛ 10 ⎞ 10
= lim
>1
⎜ ⎟ =
n→∞
n ⎝9⎠
9
n →∞
= 4 >1
The series converges conditionally.
n
( n + 1)(10 9)
a
lim n + 1 = lim
n
n→∞ a
n→∞
n(10 9)
n
∞
(4)
Note: The Ratio Test is inconclusive because
a
lim n + 1 = 1.
n→∞ a
n
n + 1⎛ 6 ⎞
6
>1
⎜ ⎟ =
n ⎝5⎠
5
n =1
19.
2
Therefore, by Theorem 9.14, the series converges.
Therefore, the series diverges by the Ratio Test.
18.
+ 1)
(−1)n +1(n + 2)
∑ n(n + 1)
n =1
an + 1 =
n
(n + 1)(6 5)
an + 1
= lim
n
n
→
∞
an
n(6 5)
n→∞
(n
∞
22.
n =1
lim
n2
2
Therefore, the series diverges by the Ratio Test.
n +1
Therefore, by the Ratio Test, the series converges.
∞
4n + 1 ( n + 1)
an + 1
= lim
2
n→∞
an
4n n
= lim
3
16. ∑
n = 0 n!
17.
1
<1
4
4n
n
∞
=
4n 3
∑ n2
Therefore, by the Ratio Test, the series diverges.
lim
3
n =1
a
(n + 1)! ⋅ 3n = lim n + 1 = ∞
lim n +1 = lim
n →∞ a
n →∞
n →∞
n!
3n +1
3
n
n →∞
+ 1)
n +1
Therefore, the series converges by the Ratio Test.
∞
n!
15. ∑ n
n=0 3
(n
n→∞
Therefore, the series converges by the Ratio Test.
∞
an + 1
(n + 1) 4
= lim
n
n→∞
an
n3 4
3
lim
n2
an + 1
( 3 2)
n2
= lim 2
⋅
n → ∞ n + 2n + 1
an
( 3 2) n
n +1
lim
n→∞
= lim
n→∞
3n 2
2( n + 2n + 1)
2
=
3
>1
2
Therefore, by the Ratio Test, the series diverges.
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
∞
25.
∞
n!
∑ n3n
27.
n =1
lim
n →∞
∞
∑
en
n=0
(n + 1)! ⋅ n3n = lim n = ∞
an +1
= lim
n →∞ ( n + 1)3n +1
n →∞ 3
an
n!
lim
n →∞
e n +1 ( n + 1)!
an + 1
= lim
n →∞
an
e n n!
⎛ n! ⎞
e
= lim e⎜⎜
= 0
⎟⎟ = lim
n →∞
n
→∞
n
n
1
!
+
+1
)⎠
⎝(
(2n)!
Therefore, the series converges by the Ratio Test.
n5
n =1
319
∑ n!
Therefore, by the Ratio Test, the series diverges.
26.
The Ratio and Root Tests
∞
a
(2n + 2)! ⋅ n5
lim n + 1 = lim
5
n→∞ a
n→∞
n
(n + 1) (2n)!
= lim
( 2n
+ 1)
5
n!
∑ nn
n =1
+ 2)( 2n + 1)n5
(n
n→∞
28.
lim
= ∞
n→∞
(n + 1)! (n + 1)
an + 1
= lim
n→∞
an
n! n n
n +1
n
1
⎛ n ⎞
= lim ⎜
⎟ =
n → ∞ ⎝ n + 1⎠
e
Therefore, by the Ratio Test, the series diverges.
Therefore, the series converges by the Ratio Test.
∞
29.
∑
n=0
6n
(n
+ 1)
n
6n +1 ( n + 2)
an + 1
= lim
n
→∞
n
an
6n ( n + 1)
n +1
lim
n →∞
n
n
= lim
n →∞
⎛ n + 1⎞
⎛ n + 1⎞
To find lim ⎜
⎟ : Let y = ⎜
⎟
n→∞ ⎝ n + 2 ⎠
⎝ n + 2⎠
6 ⎛ n + 1⎞
⎛1⎞
⎜
⎟ = 0⎜ ⎟ = 0.
n + 2⎝ n + 2 ⎠
⎝e⎠
n
ln ( n + 1) − ln ( n + 2)
⎛ n + 1⎞
ln y = n ln ⎜
⎟ =
1n
⎝ n + 2⎠
⎡ − n 2 ⎡⎣( n + 2) − ( n + 1)⎤⎦ ⎤
⎡1 ( n + 1) − 1 ( n + 2) ⎤
=
lim [ln y] = lim ⎢
lim
⎢
⎥ = −1
⎥
n→∞
n→∞
n→∞
−1 n 2
(n + 1)(n + 2) ⎥⎦
⎢⎣
⎣
⎦
by L'Hôpital's Rule. So, y →
1
.
e
Therefore, the series converges by the Ratio Test.
30.
(n!)2
∑
n = 0 (3n )!
∞
⎡( n + 1)!⎤⎦
(3n)! = lim
(n + 1)2
an +1
= lim ⎣
⋅
= 0
n →∞ (3n + 3)!
an
(n!)2 n →∞ (3n + 3)(3n + 2)(3n + 1)
2
lim
n →∞
Therefore, by the Ratio Test, the series converges.
∞
31.
5n
+1
∑ 2n
n=0
lim
n →∞
5n +1 ( 2n +1 + 1)
5( 2n + 1)
5(1 + 1 2n )
5
an +1
= lim
=
lim
=
lim
=
>1
n →∞
n →∞ 2 n +1 + 1
n →∞ 2 + 1 2 n
2
an
5n ( 2n + 1)
(
)
Therefore, the series diverges by the Ratio Test.
© 2010 Brooks/Cole, Cengage Learning
320
Chapter 9
(−1)n 24 n
∞
32.
Infinite Series
∑ (2n + 1)!
n=0
(2n + 1)! = lim
an + 1
24 n + 4
24
= lim
⋅
= 0
n
4
n → ∞ ( 2 n + 3)!
n → ∞ ( 2n + 3)( 2 n + 2)
an
2
lim
n→∞
Therefore, by the Ratio Test, the series converges.
(−1)n +1 n!
∑
( 2n
n=0 1 ⋅ 3 ⋅ 5
∞
33.
+ 1)
(n + 1)!
(2n + 1)(2n
an + 1
= lim
n→∞ 1 ⋅ 3 ⋅ 5
an
lim
n→∞
+ 3)
⋅
( 2n
1⋅3⋅5
+ 1)
= lim
n!
n→∞
n +1
1
=
2n + 3
2
Therefore, by the Ratio Test, the series converges.
Note: The first few terms of this series are −1 +
34.
∞
(−1)n 2 ⋅ 4 ⋅ 6
1
2!
3!
−
+
−
1⋅3 1⋅3⋅5 1⋅3⋅5⋅7
.
2n
∑ 2 ⋅ 5 ⋅ 8 (3n − 1)
n =1
2 ⋅ 4 2n( 2n + 2)
2 ⋅ 5 (3n − 1)
an + 1
2n + 2
2
= lim
⋅
= lim
=
n→∞ 2 ⋅ 5
n → ∞ 3n + 2
an
2 ⋅ 4 2n
3
(3n − 1)(3n + 2)
lim
n→∞
Therefore, by the Ratio Test, the series converges.
Note: The first few terms of this series are −
∞
35.
2
2⋅4
2⋅4⋅6
+
−
+
2
2⋅5
2⋅5⋅8
1
∞
∑ 5n
39.
n =1
lim
n
n→∞
∞
⎡1⎤
an = lim ⎢ n ⎥
n → ∞ ⎣5 ⎦
1
<1
5
=
lim
∞
40.
1n
lim
n
an
37.
⎛
n
⎞
∑ ⎜⎝ 2n + 1 ⎟⎠
n
n →∞
an = lim
⎛ 2n ⎞
∑ ⎜⎝ n + 1 ⎟⎠
n =1
lim
n →∞
n
n
⎛ 2n + 1 ⎞
= lim ⎜
⎟ = 2
n →∞ ⎝ n − 1 ⎠
n
n
n →∞
an = lim
n
n →∞
⎛ 4n + 3 ⎞
⎜
⎟
⎝ 2n − 1 ⎠
n
= lim
n →∞
4n + 3
= 2
2n − 1
Therefore, by the Root Test, the series diverges.
(−1)n
∑
n
n = 2 (ln n )
∞
n
n →∞
∞
⎛ 4n + 3 ⎞
∑ ⎜⎝ 2n − 1 ⎟⎠
lim
41.
⎛ n ⎞
⎜
⎟
⎝ 2n + 1 ⎠
n
= lim
n →∞
n
1
=
2n + 1
2
Therefore, by the Root Test, the series converges.
38.
n →∞
⎛ 2n + 1 ⎞
⎜
⎟
⎝ n −1⎠
n
n =1
lim
n
n =1
1
= lim
= 0
n→∞ n
Therefore, by the Root Test, the series converges.
∞
an = lim
Therefore, by the Root Test, the series diverges.
1
n =1
⎡1⎤
= lim ⎢ n ⎥
n→∞ ⎣n ⎦
n
n →∞
∑ nn
n→∞
n
n=2
1n
Therefore, by the Root Test, the series converges.
36.
⎛ 2n + 1 ⎞
∑ ⎜⎝ n − 1 ⎟⎠
lim
n →∞
n
an = lim
n →∞
n
(−1)n
(ln n)n
= lim
n →∞
1
= 0
ln n
Therefore, by the Root Test, the series converges.
n
an = lim
n →∞
n
⎛ 2n ⎞
⎜
⎟
⎝ n + 1⎠
n
= lim
n →∞
2n
= 2
n +1
Therefore, by the Root Test, the series diverges.
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
∞
42.
⎛ −3n ⎞
∑ ⎜⎝ 2n + 1 ⎟⎠
3n
∞
47.
n =1
lim
n
n→∞
an = lim
n
⎛ −3n ⎞
⎜
⎟
⎝ 2n + 1 ⎠
3n
lim
3
∑ (2
)
n +1
n =1
lim
n
n→∞
∞
n→∞
n→∞
= lim
n →∞
)
n +1
n
n
n→∞
(
ln y = lim ln n n
n →∞
n
n , let y = lim
n
To find lim
(2
(
= lim 2
n→∞
n
)
n +1
n . Then
)
49.
)
=
n=0
∞
lim
n→∞
1
∑ e3 n
an = lim
n→∞
1
⎛1⎞
= lim ⎜ n ⎟
n→∞ ⎝ 3 ⎠
e 3n
(n n )
⎛n⎞
an = lim ⎜ n ⎟
n→∞ ⎝ 3 ⎠
= lim
n→∞
n
n →∞
n1 n
1
=
3
3
⎛ n ⎞
46. ∑ ⎜
⎟
n = 1 ⎝ 500 ⎠
lim
n →∞
1
ln n
ln n =
n
n
n
n
n
(ln n)
= lim
n
n →∞
n1 n
= 0
ln n
(n!)n
(n2 )
n
(−1)n +15
n =1
n
∑
(n!)
n
(n )
2
n
= lim
n →∞
n!
= ∞
n2
5
5
<
= an
n +1
n
5
lim
= 0
n→∞ n
∞
100
1
= 100∑
n =1 n
n =1 n
∑
This is the divergent harmonic series.
∞
n
an = lim
∞
52.
ln n
1n
= lim
= 0 ⇒ y →1
n→∞ 1
n
∞
n →∞
Therefore, by the Alternating Series Test, the series
converges (conditional convergence).
n→∞
lim
∞
∑
n =1
∞
an + 1 =
Note: You can use L'Hôpital's Rule to show
lim n1 n = 1:
n→∞
2
=
n →∞
Therefore, the series converges by the Root Test.
Let y = n1 n , ln y =
ln n
= 0 <1
n
= lim
Therefore, by the Root Test, the series diverges.
51.
1n
n→∞
n →∞
(n!)n
lim
1
=
3
n
∑ 3n
n =1
n
an = lim
n
n=0
∞
lim
∑
n =1
Therefore, the series converges by the Root Test.
45.
(ln n)n
∞
50.
n
n
Therefore, by the Root Test, the series converges.
1n
n
n
n=2
n →∞
Therefore, by the Root Test, the series diverges.
∞
n →∞
∞
∑
lim
lim 2 n n + 1 = 2(1) + 1 = 3.
∑ e−3n
⎛ ln n ⎞
⎜
⎟
⎝ n ⎠
n
Therefore, by the Root Test, the series converges.
So, ln y = 0, so y = e = 1 and
44.
n
n
an = lim
n
n →∞
1
ln n
1n
ln n = lim
= lim
= 0.
n →∞ n
n →∞ 1
n
(
⎛ ln n ⎞
⎟
n ⎠
n =1
∑ ⎜⎝
lim
0
n→∞
1⎞
⎛1
⎜ − 2⎟
n ⎠
⎝n
Therefore, by the Root Test, the series converges.
48.
n
n
1⎞
⎛1
= lim ⎜ − 2 ⎟ = 0 − 0 = 0 < 1
n→∞ ⎝ n
n ⎠
n
an = lim
321
n
n→∞
3
Therefore, by the Root Test, the series diverges.
43.
1⎞
⎟
n2 ⎠
an = lim
n
n→∞
27
⎛ 3n ⎞
⎛ 3⎞
= lim ⎜
⎟ = ⎜ ⎟ =
n → ∞ ⎝ 2n + 1 ⎠
8
⎝ 2⎠
n
−
n =1
n→∞
∞
⎛1
∑ ⎜⎝ n
The Ratio and Root Tests
53.
∑n
n =1
∞
3
n
1
32
n
n =1
= 3∑
This is a convergent p-series.
an = lim
n →∞
n
⎛ n ⎞
⎜
⎟
⎝ 500 ⎠
n
⎛ n ⎞
= lim ⎜
⎟ = ∞
n →∞ ⎝ 500 ⎠
Therefore, by the Root Test, the series diverges.
© 2010 Brooks/Cole, Cengage Learning
322
Chapter 9
⎛ 2π ⎞
⎟
3 ⎠
n =1
∞
54.
∑ ⎜⎝
Infinite Series
∞
n
60.
n =1
Because r =
2π
> 1, this is a divergent Geometric
3
55.
n→∞
5n
61.
n =1
5n
5
=
2n − 1
2
lim
∞
∞
n =1
+1
n =1
n ( 2n 2 + 1)
lim
1n
n→∞
= lim
n→∞
n2
1
=
> 0
2
2n + 1
2
an + 1 =
1
lim
n =1
∞
∑
n→∞
(−1)n 3n − 2
2
n =1
n
∞
10
n =1
3 n3
∑
=
∞
∑
n =1
Because r =
58.
(−1)n 3n3−2
2
n
=
∞
1⎛ 3 ⎞
∑ 9 ⎜⎝ − 2 ⎟⎠
n
n =1
3
> 1, this is a divergent geometric series.
2
∞
64.
(10n
+ 3) n 2n
1 2n
n
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ .
n=0
n2
≤
n2
1
n3 2
Therefore, the series converges by comparison with the
p-series
∞
= lim
n→∞
1
∑ n3 2 .
n =1
10n + 3
= 10
n
Therefore, the series converges by a Limit Comparison
Test with the geometric series
∞
ln ( n)
ln ( n)
10n + 3
n 2n
lim
∑
n =1
1
n→∞
(n + 1)7 n +1 ⋅ n! = lim 7 = 0
an + 1
= lim
n→∞
an
(n + 1)! n7 n n → ∞ n
Therefore, by the Ratio Test, the series converges.
n =1
∑
n7 n
n = 1 n!
∑
n→∞
∑ n3 2 .
n =1
1
= 0
n ln (n)
lim
Therefore, the series converges by a Limit Comparison
Test with the p-series
59.
1
1
≤
= an
+ 1)ln ( n + 1)
n ln ( n)
∞
63.
10 3n3 2
10
=
n → ∞ 1 n3 2
3
∞
(n
Therefore, by the Alternating Series Test, the series
converges.
lim
∞
∑ n ln n
n=2
∑ n.
57.
cos n
converges
3n
(−1)n
∞
62.
This series diverges by limit comparison to the divergent
harmonic series
∞
∑
by Direct comparison with the convergent geometric
∞
1
cos n
series ∑ n . So, ∑ n converges.
3
n =1 3
n
∑ 2n 2
cos n
1
≤ n
n
3
3
Therefore the series
Therefore, the series diverges by the nth-Term Test
56.
2
Therefore, the series diverges by the nth-Term Test.
∑ 2n − 1
n→∞
(ln 2)2n = lim (ln 2) 2n = ∞
2n
= lim
n→∞
n→∞
4n − 1
8n
8
2
lim
Series.
∞
2n
−1
∑ 4n 2
65.
∞
(−1)n 3n −1
n =1
n!
∑
lim
n→∞
an + 1
3n
n!
3
= lim
⋅ n −1 = lim
= 0
n → ∞ ( n + 1)!
n→∞ n + 1
an
3
Therefore, by the Ratio Test, the series converges.
(Absolutely)
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
∞
66.
∑
The Ratio and Root Tests
323
(−1)n 3n
n 2n
n =1
lim
n →∞
an +1
3n +1
n2n
3n
3
= lim
⋅ n = lim
=
n
+
1
n →∞ ( n + 1)2
n →∞ 2( n + 1)
an
3
2
Therefore, by the Ratio Test, the series diverges.
67.
(−3)n
∞
∑ 3 ⋅ 5 ⋅ 7 (2n + 1)
n =1
lim
n→∞
(−3)
( 2n
an + 1
= lim
n→∞ 3 ⋅ 5 ⋅ 7
an
n +1
+ 1)( 2n + 3)
⋅
3⋅5⋅7
(−3)
( 2n
+ 1)
n
= lim
n→∞
3
= 0
2n + 3
Therefore, by the Ratio Test, the series converges.
68.
∞
∑
(2n + 1)
− 1)n!
3⋅5⋅7
n =1
lim
n→∞
18 ( 2n
n
3 ⋅ 5 ⋅ 7 ( 2n + 1)( 2n + 3)
18n ( 2n − 1)n!
(2n + 3)(2n − 1) = 1
an + 1
= lim
⋅
= lim
+
1
n
n→∞
n → ∞ 18( 2n + 1)( 2n − 1)
an
18 ( 2n + 1)( 2n − 1)n!
3 ⋅ 5 ⋅ 7 ( 2n + 1)
18
Therefore, by the Ratio Test, the series converge.
74. Replace n with n + 2.
69. (a) and (c) are the same.
∞
n5
=
n = 1 n!
∑
(n
∞
n
+ 1)5
n +1
∞
∑ (n + 1)!
n=0
= 5+
( 2)(5)
+
(3)(5)
3!
3
+
( 4)(5)
4
+
4!
75. (a) Because
310
≈ 1.59 × 10−5 ,
2 10!
10
70. (b) and (c) are the same.
∞
∑ (n + 1)( 34 )
n
=
n=0
∞
∑ n( 34 )
use 9 terms.
n −1
n =1
9
( 34 ) + 3( 34 )
=1+ 2
2
( 34 )
+ 4
( −1)
n
∑ (2n + 1)!
=
n=0
(−1)
∞
3
+
(b)
k =1
n −1
∑ (2n − 1)!
(b)
( −1)
∑ (n − 1)2n −1
n=2
n
=
∞
∑
n =1
(−1)
n +1
73. Replace n with n + 1.
n
∑ 7n =
n =1
∞
n +1
∑ 7n +1
n=0
≈ −0.7769
∞
(−3)
k
∑ 1 ⋅ 3 ⋅ 5 … ( 2k
+ 1)
=
∞
∞
n→∞
k
2 k k!
+ 1)
(−6)k k!
+ 1)!
∑ ( 2k
k =0
77. lim
(−3)
∑ (2k )!(2k
k =0
=
n2n
1
1
1
=
−
+
−
2
2 ⋅ 22
3 ⋅ 23
∞
2 k k!
1
1
+
−
3! 5!
72. (a) and (b) are the same.
∞
(−3)k
k =0
n =1
=1−
∑
76. (a) Use 10 terms, k = 9, see Exercise 3.
71. (a) and (b) are the same.
∞
9n + 2
n = 0 n!
∞
∑
=
n=2
2
2!
9n
∑ (n − 2)!
≈ 0.40967
(4n − 1) (3n + 2)an
an + 1
= lim
n→∞
an
an
= lim
n→∞
4n − 1
4
=
>1
3n + 2
3
The series diverges by the Ratio Test.
78. lim
n→∞
(2n + 1) (5n − 4)an
an + 1
= lim
n
→
∞
an
an
= lim
n→∞
2n + 1
2
=
<1
5n − 4
5
The series converges by the Ratio Test.
© 2010 Brooks/Cole, Cengage Learning
324
Chapter 9
79. lim
n→∞
Infinite Series
( n )a
(sin n + 1)
an + 1
= lim
n→∞
an
an
= lim
n→∞
n
sin n + 1
= 0 <1
n
n→∞
(cos n + 1) (n)an
an + 1
= lim
n
→
∞
an
an
cos n + 1
= 0 <1
n
The series converges by the Ratio Test.
n→∞
n →∞
(1 + (1) (n))an = lim ⎛1 + 1 ⎞ = 1
an + 1
= lim
⎜
⎟
n →∞
n →∞ ⎝
an
an
n⎠
But, lim an ≠ 0, so the series diverges.
n→∞
82. The series diverges because lim an ≠ 0.
n→∞
1
4
a3
( 14 )
= ( 12 )
13
=
a
83. lim n + 1
n→∞ a
n
≈ 0.7937
n→∞
= lim
n→∞
87. lim
2n + 1
= 0 <1
(2n)(2n + 1)
n →∞
= lim
n →∞
x
x
=
3
3
x
< 1 ⇒ −3 < x < 3.
3
For the series to converge:
n
an = lim
x +1
4
n
n →∞
n
= lim
n →∞
x +1
x +1
=
4
4
For the series to converge,
x +1
< 1 ⇒ −4 < x + 1 < 4
4
⇒ −5 < x < 3.
∞
∑ (1)
n
, diverges.
n=0
For x = −5,
∞
∑ (−1) , diverges.
n
n=0
( x + 1) (n + 1)
an + 1
= lim
n
→
∞
an
xn n
n +1
an = lim
n
n +1
= lim
n→∞
3n
y = lim
n
n +1
n→∞
n→∞
(
ln y = lim ln n n + 1
n→∞
n
n +1
3
89. lim
n→∞
= lim
n→∞
)
1
= lim ln ( n + 1)
n→∞ n
ln ( n + 1)
1
= lim
=
= 0.
n→∞
n
n +1
Because ln y = 0, y = e = 1, so
0
lim
n→∞
2( x 3)
an +1
= lim
n →∞ 2 x 3 n
an
( )
For x = 3,
n +1
1
=
<1
2n + 1
2
n +1
84. ∑ n
n=0 3
n→∞
(ln n)
1
= 0
ln n
= lim
n
The series converges by the Ratio Test.
∞
Let
1
n
1 ⋅ 3 ⋅ 5 ( 2n − 1)( 2n + 1)
1 ⋅ 2 ⋅ 3 ( 2n − 1)( 2n)( 2n + 1)
an + 1
= lim
86. lim
n→∞ a
n→∞
1 ⋅ 3 ⋅ 5 ( 2n − 1)
n
1 ⋅ 2 ⋅ 3 ( 2n − 1)
n →∞
The series converges by the Ratio Test.
n→∞
n→∞
Therefore, by the Root Test, the series converges.
88. lim
1 ⋅ 2 n( n + 1)
1 ⋅ 3 ( 2n − 1)( 2n + 1)
= lim
1⋅ 2 n
n→∞
1 ⋅ 3 ( 2n − 1)
= lim
n
an = lim
For x = −3, the series diverges.
1
2
In general, an + 1 > an > 0.
lim
n
For x = 3, the series diverges.
12
a2 =
(ln n)n
n +1
The Ratio Test is inconclusive.
a1 =
1
n=3
lim
= lim
81. lim
∞
∑
n→∞
The series converges by the Ratio Test.
80. lim
85.
n +1
1
= .
3
3
n
( x + 1) = x + 1
n +1
For the series to converge,
x + 1 < 1 ⇒ −1 < x + 1 < 1
⇒ −2 < x < 0.
∞
For x = 0, ∑
(−1)n , converges.
n =1
∞
For x = −2, ∑
n =1
n
(−1) (−1)
n
n
n
=
∞
1
∑ n , diverges.
n =1
Therefore, by the Root Test, the series converges.
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
n +1
90. lim
n →∞
2 x −1
an +1
= lim
n
n
→∞
an
2 x −1
= lim x − 1 = x − 1
lim
n →∞
exist some N > 0 such that
∞
n=0
∑ Rn
n=0
∑ 2(−1) , diverges.
n
converges, you can apply the Comparison Test to
conclude that
n=0
91. lim
n→∞
an + 1
= lim
n→∞
an
+ 1)!
x
2
x
2
n
n!
= lim ( n + 1)
n→∞
an < R for all
geometric series
n
(n
n
n > N . So, for n > N an < R n and because the
∑ 2(1) , diverges.
∞
an = r < 1
and choose R such that 0 ≤ r < R < 1. There must
⇒ 0 < x < 2.
For x = 0,
n
n→∞
x − 1 < 1 ⇒ −1 < x − 1 < 1
For x = 2,
325
100. First, let
For the series to converge,
∞
The Ratio and Root Tests
n +1
∞
∑
an
n =1
∞
converges which in turn implies that
n =1
x
= ∞
2
Second, let
lim
n +1
x +1
n!
n
n→∞
The series converges only at x = 0.
x +1
a
92. lim n +1 = lim
n →∞ a
n →∞ ( n + 1)!
n
∑ an converges.
an = r > R > 1.
Then there must exist some M > 0 such that
n
x +1
= lim
= 0
n →∞ n + 1
n
an > R for infinitely many n > M . So, for
infinitely many n > M , you have an > R n > 1 which
The series converges for all x.
implies that lim an ≠ 0 which in turn implies that
n→∞
93. See Theorem 9.17, page 641.
∞
∑ an diverges.
n =1
94. See Theorem 9.18, page 644.
95. No. Let an =
The series
∞
1
.
n + 10,000
∞
101.
n =1
1
∑ n + 10,000 diverges.
∞
⎛
∑ ⎜⎝ −100 +
n =1
an +1
1
n3 2
⎛ n ⎞
= lim
⋅
= lim ⎜
⎟
32
n
→∞
n
→∞
an
1
⎝ n + 1⎠
(n + 1)
lim
n →∞
n =1
96. One example is
1
∑ n3 2
1⎞
⎟.
n⎠
∞
102.
1
∑ n1 2
an + 1
1
n1 2
= lim
⋅
1
2
n→∞ n + 1
an
1
(
)
lim
n→∞
(Ratio Test)
12
⎛ n ⎞
= lim ⎜
⎟
n → ∞ ⎝ n + 1⎠
(b) Inconclusive
(See Ratio Test)
(c) Diverges
(Ratio Test)
(d) Diverges
(Root Test)
(e) Inconclusive
(See Root Test)
(f) Diverges
(Root Test, e > 1 )
∞
103.
1
∑ n4
4
lim
n→∞
lim an + 1 an = L > 1 or that lim an + 1 an = ∞.
n→∞
Then there exists N > 0 such that an + 1 an > 1 for all
=1
n =1
99. Assume that
n→∞
=1
n =1
97. The series converges absolutely. See Theorem 9.17.
98. (a) Converges
32
∞
104.
an + 1
1
n4
⎛ n ⎞
= lim
⋅
= lim ⎜
⎟ =1
4
n→∞ n + 1
n → ∞ ⎝ n + 1⎠
an
1
(
)
1
∑ np
n =1
n > N . Therefore,
an +1 > an , n > N ⇒ lim an ≠ 0 ⇒
n →∞
∑an diverges.
lim
n→∞
an + 1
1
np
⎛ n ⎞
= lim
⋅
= lim ⎜
⎟
p
n→∞ n + 1
n → ∞ ⎝ n + 1⎠
an
(
) 1
p
=1
© 2010 Brooks/Cole, Cengage Learning
326
Chapter 9
105.
∑ n p , p-series
∞
Infinite Series
1
n =1
lim
an = lim
n
n→∞
n
n→∞
1
1
= lim p n = 1
n→∞ n
np
So, the Root Test is inconclusive.
Note: lim n p
= 1 because if y = n p n , then
n
n→∞
p
p
ln n and
ln n → 0 as n → ∞.
n
n
ln y =
So y → 1 as n → ∞.
106. Ratio Test:
n(ln n)
an +1
= lim
= 1, inconclusive.
p
→∞
n
an
(n + 1)(ln(n + 1))
p
lim
n →∞
Root Test:
lim
n
n→∞
an = lim
n
n→∞
1
n(ln n)
p
= lim
n→∞
1
n1 n (ln n)
lim n1 n = 1. Furthermore, let y = (ln n)
p n
n→∞
ln y =
lim ln y = lim
107.
∞
p ln (ln n)
n
n→∞
n→∞
⇒
p
ln (ln n).
n
n→∞
So, lim
p n
1
n1 n (ln n)
p n
= lim
n→∞
p
p
= 0 ⇒ lim (ln n)
n→∞
ln ( n)(1 n)
n
= 1.
= 1, inconclusive.
(n!)2
∑ ( xn)!, x positive integer
n =1
(n!)
2
(a) x = 1:
∑
∑ n!, diverges
(b) x = 2:
∑ (2n)! converges by the Ratio Test:
=
n!
(n!)2
⎡( n + 1)!⎤⎦
lim ⎣
n → ∞ ( 2n + 2)!
2
(n!)2
(2n)!
= lim
n→∞
( 2n
(n
+ 1)
2
+ 2)( 2n + 1)
=
1
<1
4
(n!)
∑ (3n)! converges by the Ratio Test:
2
(c) x = 3:
⎡( n + 1)!⎤⎦
lim ⎣
n → ∞ (3n + 3)!
2
(n!)2
(3n)!
= lim
n→∞
(3n
+
(n + 1)2
3)(3n + 2)(3n
+ 1)
= 0 <1
(d) Use the Ratio Test:
⎡( n + 1)!⎤⎦
lim ⎣
n → ∞ ⎡ x( n + 1)⎤!
⎣
⎦
2
(n!)2
( xn)!
= lim ( n + 1)
n→∞
2
( xn)!
( xn + x)!
The cases x = 1, 2, 3 were solved above. For x > 3, the limit is 0. So, the series converges for all integers x ≥ 2.
© 2010 Brooks/Cole, Cengage Learning
Section 9.6
k
108. For n = 1, 2, 3, … , − an ≤ an ≤ an ⇒ − ∑ an ≤
n =1
∞
Taking limits as k → ∞, − ∑ an ≤
n =1
∞
∑ an
≤
n =1
∞
∑
k
∑ an
≤
n =1
an ⇒
n =1
The Ratio and Root Tests
327
k
∑
an .
n =1
∞
∑ an
≤
n =1
∞
∑
an .
n =1
109. The differentiation test states that if
∞
∑ Un
n =1
is an infinite series with real terms and f ( x) is a real function such that f (1 n) = U n for all positive integers n and
d 2 f dx 2 exists at x = 0, then
∞
∑ Un
n =1
converges absolutely if f (0) = f ′(0) = 0 and diverges otherwise. Below are some examples.
Convergent Series
Divergent Series
1
∑ n3 , f ( x )
1
∑ n , f ( x)
⎛
= x3
1⎞
∑ ⎜⎝1 − cos n ⎟⎠, f ( x)
= x
1
∑ sin n , f ( x)
= 1 − cos x
= sin x
110. Using the Ratio Test,
lim
n →∞
⎡ n! ⎛ 19 ⎞ n
an +1
= lim ⎢
⎜ ⎟
n →∞ ⎢ n + 1 n ⎝ 7 ⎠
an
)
⎣(
⎡
⎤
⎡ n ⋅ n n −1 ⎛ 19 ⎞⎤
1
19 1
⎛ 19 ⎞⎥
⎢
= lim ⎢
⋅ <1
⎟⎥ = nlim
⎟ =
n⎜
n⎜
n
→∞
→∞
7 e
⎢ (1 + (1 n)) ⎝ 7 ⎠⎥
⎢⎣ ( n + 1) ⎝ 7 ⎠⎥⎦
⎥⎦
⎣
⎦
(n − 1)!⎛ 19 ⎞n −1 ⎤⎥
n
n −1
⎜ ⎟
⎝7⎠
So, the series converges.
111. First prove Abel's Summation Theorem:
If the partial sums of
Let S k =
∑an are bounded and if {bn} decreases to zero, then ∑anbn converges.
k
∑ ai . Let M be a bound for { Sk }.
i =1
a1b1 + a2b2 +
+ anbn = S1b1 + ( S2 − S1 )b2 +
+ ( S n − S n −1 )bn
= S1 (b1 − b2 ) + S 2 (b2 − b3 ) +
=
n −1
∑ Si (bi
+ S n −1 (bn −1 − bn ) + S nbn
− bi + 1 ) + Snbn
i =1
The series
∞
∑ Si (bi
− bi + 1 ) is absolutely convergent because Si (bi − bi + 1 ) ≤ M (bi − bi + 1 ) and
i =1
∞
∑ (bi
− bi + 1 ) converges
i =1
to b1.
Also, lim S nbn = 0 because {Sn} bounded and bn → 0. Thus,
n→∞
Now let bn =
∞
∑ anbn
n =1
n
= lim ∑ ai bi converges.
n→∞
i =1
1
to finish the problem.
n
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
328
Infinite Series
Section 9.7 Taylor Polynomials and Approximations
1. y = − 12 x 2 + 1
7. f ( x) = sec x
Parabola
f ′( x) = sec x tan x
Matches (d)
2. y =
1 x4
8
1 x2
2
−
+1
⎛π ⎞
f⎜ ⎟ =
⎝4⎠
2
⎛π ⎞
f ′⎜ ⎟ =
⎝4⎠
2
y-axis symmetry
π⎞
⎛π ⎞
⎛ π ⎞⎛
P1 ( x) = f ⎜ ⎟ + f ′⎜ ⎟⎜ x − ⎟
4⎠
⎝4⎠
⎝ 4 ⎠⎝
Three relative extrema
P1 ( x) =
Matches (c)
π⎞
⎛
2⎜ x − ⎟
4⎠
⎝
2 +
5
3. y = e
−1 2
⎡⎣( x + 1) + 1⎤⎦
f
)
Linear
π
,
4
2
)
P1
␲
−
4
Matches (a)
␲
2
−1
4. y = e
−1 2 ⎡ 1
⎣3
(x
− 1)
3
− ( x − 1) + 1⎤
⎦
P1 is called the first degree Taylor polynomial for f at
Cubic
8
= 8 x −1 2
x
5. f ( x) =
f ′( x) = − 4 x −3 2
f ( 4) = 4
f ′( 4) = −
P1 ( x ) = f ( 4) + f ′( 4)( x − 4)
π⎞
⎛π ⎞
⎛ π ⎞⎛
P1 = f ⎜ ⎟ + f ′⎜ ⎟⎜ x − ⎟ = 1 +
4⎠
⎝4⎠
⎝ 4 ⎠⎝
P1 ( x) = 2 x + 1 −
1
1
= 4 − ( x − 4) = − x + 6
2
2
.
⎛π ⎞
f ′⎜ ⎟ = 2
⎝4⎠
f ′( x) = sec 2 x
1
2
4
⎛π ⎞
f⎜ ⎟ = 1
⎝4⎠
8. f ( x) = tan x
Matches (b)
π
π⎞
⎛
2⎜ x − ⎟
4⎠
⎝
π
2
3
8
f
−␲
2
␲
2
(4, 4)
P1
−3
−1
11
P1 is called the first degree Taylor polynomial for f at
0
P1 is the first degree Taylor polynomial for f at 4.
6. f ( x) =
3
6
= 6 x −1 3
x
f (8) = 3
f ′(8) = −
f ′( x) = −2 x − 4 3
1
8
P1 ( x) = f (8) + f ′(8)( x − 8)
= 3−
1
1
( x − 8) = − x + 4
8
8
4
(8, 3)
f
P1
12
0
2
P1 is the first degree Taylor polynomial for f at 8.
© 2010 Brooks/Cole, Cengage Learning
π
4
.
Section 9.7
9. f ( x) =
4
= 4 x −1 2
x
Taylor Polynomials and Approximations
329
f (1) = 4
f ′( x) = −2 x −3 2
f ′(1) = −2
f ′′( x) = 3 x
f ′′(1) = 3
−5 2
P2 = f (1) + f ′(1)( x − 1) +
f ′′(1)
2
(x
− 1)
2
3
2
( x − 1)
2
= 4 − 2( x − 1) +
10
P2
(1, 4)
f
−2
6
−2
x
0
0.8
0.9
1.0
1.1
1.2
2
f ( x)
Error
4.4721
4.2164
4.0
3.8139
3.6515
2.8284
P2 ( x)
7.5
4.46
4.215
4.0
3.815
3.66
3.5
⎛π ⎞
f⎜ ⎟ =
⎝4⎠
2
f ′( x) = sec x tan x
⎛π ⎞
f ′⎜ ⎟ =
⎝4⎠
2
f ′′( x) = sec3 x + sec x tan 2 x
⎛π ⎞
f ′′⎜ ⎟ = 3 2
⎝4⎠
10. f ( x) = sec x
2
f ′′(π 4) ⎛
π⎞
π⎞
⎛π ⎞
⎛ π ⎞⎛
P2 ( x) = f ⎜ ⎟ + f ′⎜ ⎟⎜ x − ⎟ +
⎜x − ⎟
4⎠
2 ⎝
4⎠
⎝4⎠
⎝ 4 ⎠⎝
P2 ( x) =
2 +
π⎞ 3
⎛
2⎜ x − ⎟ +
4⎠ 2
⎝
π⎞
⎛
2⎜ x − ⎟
4⎠
⎝
2
4
−3
3
0
–2.15
0.585
0.685
π 4
0.885
0.985
1.785
f ( x)
–1.8270
1.1995
1.2913
1.4142
1.5791
1.8088
– 4.7043
P2 ( x)
15.5414
1.2160
1.2936
1.4142
1.5761
1.7810
4.9475
x
© 2010 Brooks/Cole, Cengage Learning
330
11.
Chapter 9
Infinite Series
f ( x) = cos x
12. f ( x) = x 2e x , f (0) = 0
P2 ( x) = 1 −
1 x2
2
P4 ( x) = 1 −
1 2
x
2
P6 ( x) = 1 −
1 x2
2
(a)
+
+
1 x4
24
1 x6
720
−
2
P6
−3
P4
f ′( x) = −sin x
P2′ ( x) = − x
f ′′( x) = −cos x
P2′′ ( x) = −1
f ′′′( x) = ( x 2 + 6 x + 6)e x
f ′′′(0) = 6
f (4) ( x) = ( x 2 + 8 x + 12)e x
f (4) (0) = 12
(b)
f ′′′( x) = sin x
P4′′′ ( x) = x
f (4) ( x) = cos x
P4(4) ( x) = 1
P6(5) ( x) = − x
f (6) ( x) = −cos x
P ( 6 ) ( x ) = −1
P2
f
−3
3
−1
(c)
f ′′(0) = 2 = P2′′ (0)
f ′′′(0) = 6 = P3′′′ (0)
f (4) (0) = 12 = P4(4) (0)
(c) In general, f (n) (0) = Pn (n) (0) for all n.
(d) f (n) (0) = Pn(n) (0)
f (0) = 1
f ′( x) = 3e3 x
f ′(0) = 3
f ′′(0) = 9
3x
3x
f ′′′(0) = 27
f (4) ( x) = 81e3 x
f (4) (0) = 81
P4 ( x) = f (0) + f ′(0) x +
= 1 + 3x +
f ′′(0)
2!
f ′′′(0)
x2 +
3!
x3 +
f ( 4 ) ( 0)
4!
x4
9 2
9
27 4
x + x3 +
x
2
2
8
f ( x) = e − x
f ′( x) = −e
x4
12 x 4
= x 2 + x3 +
4!
2
2
P4
f (6) (0) = −1 = P6(6) (0)
f ′′′( x) = 27e
6 x3
= x 2 + x3
3!
P3
f (5) ( x) = −sin x
f ( x ) = e3 x
2x2
= x2
2!
P4 ( x) = x 2 + x3 +
f (4) (0) = 1 = P4(4) (0)
14.
f ′′(0) = 2
P3 ( x) = x 2 +
f ′′(0) = P2′′ (0) = −1
f ′′( x) = 9e
f ′′( x) = ( x 2 + 4 x + 2)e x
P2
−2
13.
f ′(0) = 0
P2 ( x) =
3
f
(b)
f ′( x) = ( x 2 + 2 x)e x
(a)
1 4
x
24
f (0) = 1
f ′(0) = −1
−x
f ′′( x) = e − x
f ′′(0) = 1
f ′′′( x) = −e − x
f ′′′(0) = −1
f ( 4) ( x ) = e − x
f (4) (0) = 1
f (5) ( x) = −e − x
f (5) (0) = −1
P5 ( x) = f (0) + f ′(0) x +
+
f (5) (0)
5!
f ′(0)
2!
x2 +
x5 = 1 − x +
f ′′′(0)
3!
x3 +
f (4) (0)
4!
x4
x2
x3
x4
x5
−
+
−
2
6
24 120
© 2010 Brooks/Cole, Cengage Learning
Section 9.7
f ( x) = e − x 2
15.
f ′( x) =
f ′′( x) =
f ′′′( x) =
f ( 4) ( x ) =
f ′′(0) =
f ′′( x) =
f ′′′( x) =
f
( 4)
( x)
=
1
4
1
8
1
f (4) (0) =
16
f ′′′(0) = −
f ′′(0)
2!
x2 +
f ′′′(0)
3!
x3 +
f (4) (0)
4!
x4
1
1
1 3
1 4
x + x2 −
x +
x
2
8
48
384
f ( x) = e x 3
f ′( x) =
1
2
f ′(0) = −
P4 ( x) = f (0) + f ′(0) x +
16.
f (0) = 1
1 x3
e
3
1 x3
e
9
1 x3
e
27
1 x3
e
81
f ′(0) =
f ′′(0) =
f ′′′(0) =
f
(4)
( x)
=
19.
1
1 9 2 1 27 3 1 81 4
x +
x +
x +
x
3
2!
3!
4!
1
1 2
1 3
1 4
=1+ x +
x +
x +
x
3
18
162
1944
f ( x) = sin x
f ′′′(0) = −1
f (4) ( x) = sin x
f ( 4 ) ( 0) = 0
f (5) ( x) = cos x
f (5) (0) = 1
0 2
0
1
−1 3
x +
x + x 4 + x5
2!
3!
4!
5!
1
1 5
x
= x − x3 +
6
120
P5 ( x) = 0 + (1) x +
18.
f ( x) = sin π x
f ( 0) = 0
f ′( x) = π cos π x
f ′′( x) = −π 2 sin π x
f ′′′( x) = −π 3 cos π x
P3 ( x) = 0 + π x +
f ′(0) = π
f ′′(0) = 0
f ′′′(0) = −π 3
0 2
−π 3 3
π3 3
x +
x = πx −
x
2!
3!
6
f ′′′(0) = 3
f (4) ( x) = xe x + 4e x
f (4) (0) = 4
2 2
3
4
x + x3 + x 4
2!
3!
4!
1
1
= x + x 2 + x3 + x 4
2
6
20.
f ( x) = x 2e − x
f (0) = 0
f ′( x) = 2 xe − x − x 2e − x
f ′′( x) = 2e
−x
f ′′′( x) = −6e
− 4 xe
−x
−x
+ 6 xe
f ′(0) = 0
f ′′(0) = 2
2 −x
+ xe
−x
f ′′′(0) = −6
2 −x
− xe
f (4) ( x) = 12e − x − 8 xe − x + x 2e − x
f ′′(0) = 0
f ′′′( x) = −cos x
f ′′(0) = 2
f ′′′( x) = xe x + 3e x
P4 ( x) = 0 + x +
f ′(0) = 1
f ′′( x) = −sin x
f (0) = 0
f ′(0) = 1
f ′′( x) = xe x + 2e x
f ( 0) = 0
f ′( x) = cos x
f ( x) = xe x
f ′( x) = xe x + e x
1
3
1
9
1
27
1
81
P4 ( x) = 1 +
17.
331
f ( 0) = 1
1
− e−x 2
2
1 −x 2
e
4
1
− e−x 2
8
1 −x 2
e
16
=1−
Taylor Polynomials and Approximations
f (4) (0) = 12
2 2
−6 3 12 4
x +
x +
x
2!
3!
4!
1
= x 2 − x3 + x 4
2
P4 ( x) = 0 + 0 x +
21.
f ( x) =
1
−1
= ( x + 1)
x +1
f ′( x) = −( x + 1)
f ′′( x) = 2( x + 1)
f ′(0) = −1
−2
f ′′(0) = 2
−3
f ′′′( x) = −6( x + 1)
f (4) ( x) = 24( x + 1)
−4
f ′′′(0) = −6
−5
f (4) (0) = 24
f (5) ( x) = −120( x + 1)
P5 ( x) = 1 − x +
f (0) = 1
−6
f (5) (0) = −120
2 x2
6 x3
24 x 4 120 x5
−
+
−
2!
3!
4!
5!
= 1 − x + x 2 − x3 + x 4 − x5
© 2010 Brooks/Cole, Cengage Learning
332
Chapter 9
f ( x) =
22.
Infinite Series
x
x +1−1
=
x +1
x +1
= 1 − ( x + 1)
f ′( x) = ( x + 1)
f ′′′( x) = 6( x + 1)
f
( 4)
( x)
f ′′′( x) = 4 sec 2 x tan 2 x + 2 sec4 x
f ′′(0) = −2
P3 ( x) = 0 + 1( x) + 0 +
f (4) (0) = −24
−5
2 2
6
24 4
x + x3 −
x
2
6
24
= x − x 2 + x3 − x 4
23.
f ( x) = sec x
f ( 0) = 1
f ′(0) = 0
f ′′( x) = sec3 x + sec x tan 2 x
26.
= x + 13 x3
f (1) = 2
f ′(1) = −2
f ′′( x) = 4 x −3
f ′′(1) = 4
f ′′′(1) = −12
P3 ( x) = 2 − 2( x − 1) +
f ′′(0) = 1
2 x3
6
f ′′(0) = 0
f ′′′(0) = 2
4
12
( x − 1)2 − ( x − 1)3
2!
3!
= 2 − 2( x − 1) + 2( x − 1) − 2( x − 1)
2
3
1 2
1
x = 1 + x2
2!
2
1
= x −2
x2
f ′( x) = −2 x −3
f ( x) =
f ′′( x) = 6 x
2
= 2 x −1
x
f ′( x) = −2 x −2
f ( x) =
25.
f ′′′( x) = −12 x −4
f ′( x) = sec x tan x
P2 ( x) = 1 + 0 x +
f ′(0) = 1
f ′(0) = 1
f ′′′(0) = 6
−4
= −24( x + 1)
f (0) = 0
f ′( x) = sec 2 x
f ′′( x) = 2 sec 2 x tan x
−3
P4 ( x) = 0 + 1( x) −
f ( x) = tan x
24.
−1
−2
f ′′( x) = −2( x + 1)
f (0) = 0
−4
f ( 2) = 1 4
f ′( 2) = −1 4
f ′′( 2) = 3 8
f ′′′( x) = −24 x −5
f ′′′( 2) = −3 4
f (4) ( x) = 120 x −6
f (4) ( 2) = 15 8
1
1
38
34
15 8
2
3
4
− ( x − 2) +
( x − 2) − ( x − 2) +
( x − 2)
4
4
2!
3!
4!
1
1
3
1
5
2
3
=
− ( x − 2) + ( x − 2) − ( x − 2) +
( x − 2) 4
4
4
16
8
64
P4 ( x) =
27.
f ( x) =
x = x1 2
1 −1 2
x
2
1
f ′′( x) = − x −3 2
4
3 −5 2
f ′′′( x) = x
8
f ′( x) =
f ( 4) = 2
f ′( 4) =
1
4
1
32
3
f ′′′( 4) =
256
f ′′( 4) = −
1
1 32
3 256
2
3
( x − 4) −
( x − 4) +
( x − 4)
4
2!
3!
1
1
1
= 2 + ( x − 4) −
( x − 4) 2 +
( x − 4) 3
4
64
512
P3 ( x) = 2 +
28.
f ( x) = x1 3
f (8) = 2
1
f ′( x) = x −2 3
3
2
f ′′( x) = − x −5 3
9
10 −8 3
f ′′′( x) =
x
27
P3 ( x) = 2 +
f ′(8) =
1
12
1
144
10 1
5
⋅
=
f ′′′(8) =
27 28
3456
f ′′(8) = −
1
1
5
( x − 8) −
( x − 8)2 +
( x − 8)3
12
288
20,736
© 2010 Brooks/Cole, Cengage Learning
Section 9.7
f ( x) = ln x
29.
333
f ( 2) = ln 2
1
= x −1
x
f ′′( x) = − x −2
f ′( x) =
f ′′′( x) = 2 x
Taylor Polynomials and Approximations
f ′( 2) = 1 2
f ′′( 2) = −1 4
f ′′′( 2) = 1 4
−3
f (4) ( x) = −6 x −4
f (4) ( 2) = −3 8
1
14
14
38
2
3
4
( x − 2) − ( x − 2) + ( x − 2) − ( x − 2)
2
2!
3!
4!
1
1
1
1
2
= ln 2 + ( x − 2) − ( x − 2) +
( x − 2) 3 − ( x − 2) 4
2
8
24
64
P4 ( x) = ln 2 +
30.
f ( x) = x 2 cos x
f (π ) = −π 2
f ′( x) = cos x − x sin x
f ′(π ) = −2π
2
f ′′( x) = 2 cos x − 4 x sin x − x cos x
f ′′(π ) = −2 + π 2
2
P2 ( x) = −π 2 − 2π ( x − π ) +
31. (a) P3 ( x) = π x +
π3
(π 2
− 2)
2
(x
− π)
2
x3
3
2
1⎞
1⎞
8 ⎛
1⎞
⎛
⎛
(b) Q3 ( x) = 1 + 2π ⎜ x − ⎟ + 2π 2 ⎜ x − ⎟ + π 3 ⎜ x − ⎟
4⎠
4⎠
3 ⎝
4⎠
⎝
⎝
3
4
−0.5
0.5
P3
f
Q3
−4
−2 2
0
24 4
x + x3 +
x = 1 − x2 + x4
2!
3!
4!
32. (a) P4 ( x) = 1 + 0 x +
2
P4
f
−3
3
Q4
P2
−2
(b) Q4 ( x) =
1 ⎛ 1⎞
12
0
−3
1
1
1
1
2
3
4
2
4
+ ⎜ − ⎟( x − 1) +
( x − 1) + ( x − 1) + ( x − 1) = − ( x − 1) + ( x − 1) − ( x − 1)
2 ⎝ 2⎠
2!
3!
4!
2
2
4
8
33. f ( x) = sin x
P1 ( x) = x
P3 ( x) = x −
1 x3
6
P5 ( x) = x −
1 x3
6
(a)
+
1 x5
120
x
0.00
0.25
0.50
0.75
1.00
sin x
0.0000
0.2474
0.4794
0.6816
0.8415
P1 ( x)
0.0000
0.2500
0.5000
0.7500
1.0000
P3 ( x)
0.0000
0.2474
0.4792
0.6797
0.8333
P5 ( x)
0.0000
0.2474
0.4794
0.6817
0.8417
© 2010 Brooks/Cole, Cengage Learning
334
Chapter 9
Infinite Series
(b)
3
P3
P1
f
− 2␲
2␲
P5
−3
(c) As the distance increases, the accuracy decreases.
34. (a)
f ( x) = e x
f (1) = e
f ′( x) = e x
f ′(1) = e
f ′′( x) = f ′′′( x) = f (4) ( x) = e x and f ′′(1) = f ′′′(1) = f (4) (1) = e
P1 ( x) = e + e( x − 1)
e
( x − 1)2
2
e
e
e
2
3
P4 ( x) = e + e( x − 1) + ( x − 1) + ( x − 1) +
( x − 1)4
2
6
24
P2 ( x) = e + e( x − 1) +
x
1.00
1.25
1.50
1.75
2.00
ex
e
3.4903
4.4817
5.7546
7.3891
P1 ( x)
e
3.3979
4.0774
4.7570
5.4366
P2 ( x)
e
3.4828
4.4172
5.5215
6.7957
P4 ( x)
e
3.4903
4.4809
5.7485
7.3620
(b)
7
P4
P1
P2
y = ex
−6
6
−1
(c) As the degree increases, the accuracy increases. As the distance from x to 1 increases, the accuracy decreases.
35. f ( x) = arcsin x
(a) P3 ( x) = x +
(b)
x
x3
6
–0.75
–0.50
–0.25
0
0.25
0.50
0.75
f ( x)
–0.848
–0.524
–0.253
0
0.253
0.524
0.848
P3 ( x)
–0.820
–0.521
–0.253
0
0.253
0.521
0.820
(c)
y
π
2
f
x
−1
1
P3
−
π
2
© 2010 Brooks/Cole, Cengage Learning
Section 9.7
Taylor Polynomials and Approximations
335
36. (a) f ( x) = arctan x
x3
3
P3 ( x) = x −
(b)
x
–0.75
–0.50
–0.25
0
0.25
0.50
0.75
f ( x)
–0.6435
–0.4636
–0.2450
0
0.2450
0.4636
0.6435
P3 ( x)
–0.6094
–0.4583
–0.2448
0
0.2448
0.4583
0.6094
(c)
y
π
2
π
4
P3
x
−1
1
1
2
−π
4
f
−π
2
39. f ( x) = ln ( x 2 + 1)
37. f ( x) = cos x
P8
y
P4
y
6
P6
P2
3
4
2
f(x) = cos x
2
f (x) = ln (x 2 + 1)
x
−6
6
x
−4 −3 −2
8
2
−4
−6
−3
P6 P2
40. f ( x) = 4 xe − x
38. f ( x) = arctan x
y
y
P7
P8
P5 P1
3
4
P4
2 4
2
y = 4xe(−x /4)
4
2
2
1
f (x) = arctan x
x
−4
x
−3 −2
1
4
3
−2
P9 P13
P11 P3
41. f ( x) = e3 x ≈ 1 + 3x +
f
( 12 ) ≈ 4.3984
9 2
x
2
42. f ( x) = x 2e− x ≈ x 2 − x3 +
f
43.
( 15 ) ≈ 0.0328
f ( x) = ln x ≈ ln ( 2) +
1
2
+
9 3
x
2
+
27 4
x
8
1 4
x
2
(x
− 2) −
1
8
(x
− 2) +
2
1
24
(x
− 2) −
3
1
64
(x
− 2)
4
f ( 2.1) ≈ 0.7419
44.
⎛π 2 − 2⎞
2
f ( x) = x 2 cos x ≈ −π 2 − 2π ( x − π ) + ⎜
⎟( x − π )
⎝ 2 ⎠
⎛ 7π ⎞
f ⎜ ⎟ ≈ −6.7954
⎝ 8 ⎠
© 2010 Brooks/Cole, Cengage Learning
336
Chapter 9
Infinite Series
45. f ( x) = cos x; f (5) ( x) = −sin x ⇒ Max on [0, 0.3] is 1.
R4 ( x) ≤
1
(0.3)5 = 2.025 × 10−5
5!
Note: you could use R5 ( x): f (6) ( x) = −cos x, max on [0, 0.3] is 1.
R5 ( x) ≤
1
6
(0.3) = 1.0125 × 10−6
6!
Exact error: 0.000001 = 1.0 × 10−6
46. f ( x) = e x ; f (6) ( x) = e x ⇒ Max on [0, 1] is e1.
R5 ( x) ≤
e1 6
(1) ≈ 0.00378 = 3.78 × 10−3
6!
x(6 x 2 + 9)
⇒ Max on [0, 0.4] is f (4) (0.4) ≈ 7.3340.
47. f ( x) = arcsin x; f (4) ( x) =
72
2
(1 − x )
R3 ( x) ≤
7.3340
4
(0.4) ≈ 0.00782 = 7.82 × 10−3. The exact error is 8.5 × 10−4. [Note: You could use R4 .]
4!
48. f ( x) = arctan x; f (4) ( x) =
24 x( x 2 + 1)
(1 − x 2 )
4
⇒ Max on [0, 0.4] is f (4) (0.4) ≈ 22.3672.
22.3672
4
R3 ( x) ≤
(0.4) ≈ 0.0239
4!
52. f ( x) = ln x, f ′( x) = x −1 , f ′′( x) = − x −2 , …
f (n + 1) ( x) = ( −1)
Rn ≤
(0.25)n +1
By trial and error, n = 3
53. f ( x) = ln ( x + 1)
f (n + 1) ( x) =
f (n + 1) ( x) ≤ 1 for all x and all n.
Rn ( x) =
f (n + 1) ( z ) x n + 1
(n
+ 1)!
(0.1)
(n + 1)!
n +1
≤
< 0.001
n +1
By trial and error, n = 3.
50. f ( x) = cos x
xn +1
n!
n +1
(0.25) < 0.001
n
1
!
+
(
)
g (n + 1) ( x) ≤ 1 for all x.
1
n +1
(0.3) < 0.001
(n + 1)!
n!
The maximum value of f (n + 1) ( x) on [1, 1.25] is n!
49. g ( x) = sin x
Rn ( x) ≤
n
(−1)n n!
( x + 1)n +1
⇒ Max on [0, 0.5] is n!.
(0.5) < 0.0001
n!
(0.5)n +1 =
n +1
(n + 1)!
n +1
Rn ≤
By trial and error, n = 9. (See Example 9.) Using 9
terms, ln (1.5) ≈ 0.4055.
< 0.001
By trial and error, n = 2.
51.
f ( x) = e x
f (n + 1) ( x) = e x
Max on [0, 0.6] is e0.6 ≈ 1.8221.
Rn ≤
1.8221
n +1
(0.6) < 0.001
1
!
n
+
(
)
By trial and error, n = 5.
© 2010 Brooks/Cole, Cengage Learning
Section 9.7
54. f ( x) = cos(π x 2 )
58.
x2
x4
x6
+
−
+
2!
4!
6!
g ( x) = cos x = 1 −
f ( x) = g (π x 2 )
(π x )
2
=1−
=1−
f (0.6) = 1 −
(π x )
2
2
+
2!
π 2 x4
+
2!
π2
4
4!
π 4 x8
4!
(0.6)4
2!
Taylor Polynomials and Approximations
+
−
π4
4!
(π x )
2
−
−
π6
6!
π 2n
(0.6)
(2n)!
4n
−0.3936 < x < 0.3936
(0.6)12
+
f (n + 1) ( x) ≤ 1 for all x and all n.
R5 ( x) ≤
< 0.0001.
f ′( x) = ( −π )e−π x
6
6
< 0.001
< 0.72
x < 0.9467
−0.9467 < x < 0.9467
window [−0.9467, 0.9467] × [−0.001, 0.001] to verify
n + 1 −π x
e
(π ) 1.3 n +1
( )
(n + 1)!
≤ ( −π )
n +1
on [0, 1.3]
n +1
< 0.0001
the answer.
60. f ( x) = e −2 x ≈ 1 − 2 x + 2 x 2 −
4 3
x
3
By trial and error, n = 16. Using 16 terms,
f ′( x) = −2e −2 x , f ′′( x) = 4e −2 x ,
e −π (1.3) ≈ 0.01684.
f ′′′( x) = −8e −2 x , f (4) ( x) = 16e −2 x
56. f ( x) = e − x
R3 ( x) =
f ′( x) = −e − x
f (n + 1) ( x) = ( −1)
Rn ≤
(n
n +1 − x
e
≤ 1 on [0, 1]
1
1n + 1 ≤ 0.0001
+ 1)!
x2
x3
+ ,x < 0
f ( x) = e ≈ 1 + x +
2
6
x
R3 ( x) =
ez 4
x < 0.001
4!
e z x 4 < 0.024
xe z 4 < 0.3936
0.3936
< 0.3936, z < 0
ez 4
−0.3936 < x < 0
x <
f 4 ( z)
16e −2 z 4
2
4
x = e−2 z x 4 < 0.001
( x − 0) =
4!
24
3
e −2 z x 4 < 0.0015
By trial and error, n = 7.
57.
x
1
x
6!
Note: Use a graphing utility to graph
y = cos x − (1 − x 2 2 + x 4 24) in the viewing
55. f ( x) = e−π x , f (1.3)
Rn ≤
x2
x4
+
, fifth degree polynomial
2!
4!
59. f ( x) = cos x ≈ 1 −
By trial and error, n = 4. Using 4 terms
f (0.6) ≈ 0.4257.
f (n + 1) ( x) = ( −π )
x4
sin z 4
x ≤
< 0.001
4!
4!
x < 0.3936
+
Because this is an alternating series,
Rn ≤ an + 1 =
x3
3!
x 4 < 0.024
6
+
6!
(0.6)8
R3 ( x) =
6!
π 6 x12
f ( x) = sin x ≈ x −
337
14
⎛ 0.0015 ⎞
x < ⎜ −2 z ⎟
⎝ e
⎠
≈ 0.1970e 2 z < 0.1970, for z < 0.
So, 0 < x < 0.1970.
In fact, by graphing f ( x) = e −2x and
y = 1 − 2x + 2x2 −
4 3
x , you can verify that
3
f ( x) − y < 0.001 on ( −0.19294, 0.20068).
61. The graph of the approximating polynomial P and the
elementary function f both pass through the point
(c, f (c)) and the slopes of P and f agree at
(c, f (c)). Depending on the degree of P, the nth
derivatives of P and f agree at (c, f (c)).
62. f (c) = P2 (c), f ′(c) = P2′ (c), and f ′′(c) = P2′′ (c)
63. See definition on page 652.
© 2010 Brooks/Cole, Cengage Learning
338
Chapter 9
Infinite Series
64. See Theorem 9.19, page 656.
65. As the degree of the polynomial increases, the graph of
the Taylor polynomial becomes a better and better
approximation of the function within the interval of
convergence. Therefore, the accuracy is increased.
66.
10
f
P1
8
(b) R2 ( x) = −1 +
π 2 ( x + 2)
2
32
π ( x − 6)
2
2
32
(c) No. The polynomial will be linear. Horizontal
translations of the result in part (a) are possible only
at x = −2 + 8n (where n is an integer) because the
period of f is 8.
y
P2
69. (a) Q2 ( x) = −1 +
6
4
70. Let f be an odd function and Pn be the nth Maclaurin
polynomial for f. Because f is odd, f ′ is even:
2
x
−20 P
3
−2
10
20
−4
f ( − x + h) − f ( − x)
h
− f ( x − h) + f ( x)
= lim
h →0
h
f ′( − x) = lim
h →0
67. (a) f ( x) = e x
P4 ( x) = 1 + x +
1 2
1
1 4
x + x3 +
x
2
6
24
g ( x) = xe x
= lim
1 3
1
1 5
x + x4 +
x
2
6
24
Q5 ( x) = x P4 ( x)
f , f ′′, f (4) , etc. are all odd functions, which implies that
Q6 ( x) = x P5 ( x) = x 2 −
= 0. So, in the formula
Pn ( x) = f (0) + f ′(0) x +
x3
x5
P5 ( x) = x −
+
3!
5!
g ( x) = x sin x
= f ′( x).
Similarly, f ′′ is odd, f ′′′ is even, etc. Therefore,
f (0) = f ′′(0) =
(b) f ( x) = sin x
f ′′(0) x 2
+ all the
2!
coefficients of the even power of x are zero.
x4
x6
+
3!
5!
sin x
1
x2
x4
= P5 ( x) = 1 −
+
x
x
3!
5!
68. (a) P5 ( x) = x −
−h
h →0
Q5 ( x) = x + x 2 +
(c) g ( x) =
f ( x + ( − h)) − f ( x)
x3
x5
for f ( x) = sin x
+
3!
5!
71. Let f be an even function and Pn be the nth Maclaurin
polynomial for f. Because f is even, f ′ is odd, f ′′ is
even, f ′′′ is odd, etc. All of the odd derivatives of f are
odd and so, all of the odd powers of x will have
coefficients of zero. Pn will only have terms with even
powers of x.
72. Let
x2
x4
P5′ ( x) = 1 −
+
2!
4!
Pn ( x) = a0 + a1 ( x − c) + a2 ( x − c) +
This is the Maclaurin polynomial of degree 4 for
g ( x) = cos x.
where ai =
x2
x4
x6
(b) Q6 ( x) = 1 −
for cos x
+
−
2
4!
6!
x3
x5
Q6′ ( x ) = − x +
−
= − P5 ( x)
3!
5!
(c) R( x) = 1 + x +
R′( x) = 1 + x +
x2
x3
x4
+
+
2!
3!
4!
2
f (i ) ( c )
i!
+ an ( x − c)
.
Pn (c) = a0 = f (c)
For
1 ≤ k ≤ n,
⎛ f (k ) (c) ⎞
Pn(k ) (c) = an k! = ⎜
⎟ k! = f (k ) (c).
⎜ k! ⎟
⎝
⎠
73. As you move away from x = c, the Taylor Polynomial
becomes less and less accurate.
x2
x3
+
2!
3!
The first four terms are the same!
© 2010 Brooks/Cole, Cengage Learning
n
Section 9.8
Power Series
339
Section 9.8 Power Series
1. Centered at 0
9.
∞
x 2n
∑ (2n)!
2. Centered at 0
n=0
3. Centered at 2
L = lim
n→∞
4. Centered at π
5.
∞
∑ (−1)
n
n=0
(−1) x
un + 1
= lim
n→∞
un
n + 2
n +1 n +1
n→∞
n→∞
= lim
⋅
n→∞
n +1
( −1) x n
n
n +1
x = x
n + 2
10.
∞
x2
= 0
(2n + 2)(2n + 1)
∞
(2n)!x 2n
n=0
n!
∑
L = lim
∑ ( 4 x)
x 2 n ( 2n)!
So, the series converges for all x ⇒ R = ∞.
x <1⇒ R =1
6.
x(2 n + 2) ( 2n + 2)!
n→∞
xn
n +1
L = lim
= lim
= lim
un + 1
un
n→∞
n
un + 1
un
n=0
( 4 x)
u
L = lim n +1 = lim
n →∞ u
n →∞
n
( 4 x) n
∞
7.
∑
n =1
( 4 x)
n
= lim
n →∞
8.
∑
n=0
(−1)
n
1
4
∞
11.
(4 x)n +1 (n + 1)2
n
( 4 x) n 2
= lim
n →∞
12.
n2
( n + 1)
2
( 4 x)
+ 2)( 2n + 1) x 2
(n
+ 1)
= ∞
n
⎛ x⎞
∑ ⎜⎝ 7 ⎟⎠
Because the series is geometric, it converges only if
x
< 1, or −7 < x < 7.
7
1
4
∞
13.
∑
( −1)
n =1
5n
n
n=0
= 4x
xn
<1⇒ R = 5
⎛ x⎞
∑⎜ ⎟
n=0 ⎝ 4 ⎠
∞
n +1
5
( 2n
n!
Because the series is geometric, it converges only if
x
< 1, or − 4 < x < 4.
4
(−1) x n +1 5n +1 = lim x = x
u
L = lim n +1 = lim
n →∞ u
n →∞
n →∞ 5
5
n
(−1)n x n 5n
x
(2n)!x 2 n
The series only converges at x = 0. R = 0.
n →∞
u n +1
un
4 x <1⇒ R =
∞
= lim 4 x = 4 x
2
n →∞
+ 2)!x 2 n + 2 ( n + 1)!
n→∞
n
L = lim
( 2n
n→∞
= lim
n +1
4 x <1⇒ R =
= lim
n
xn
n
( −1) x n +1 ⋅ n
un + 1
= lim
n→∞
un
n +1
(−1)n x n
n +1
lim
n→∞
= lim
n→∞
nx
= x
n +1
Interval: −1 < x < 1
When x = 1, the alternating series
When x = −1, the p-series
∞
∞
( −1)n
n =1
n
∑
converges.
1
∑ n diverges.
n =1
Therefore, the interval of convergence is ( −1, 1].
© 2010 Brooks/Cole, Cengage Learning
340
Chapter 9
Infinite Series
∞
14.
n +1
∑ (−1) (n + 1) x n
16.
n=0
(3 x )
n
n=0
(−1) (n + 2) x
un + 1
= lim
n
n→∞
un
(−1) (n + 1) x n
n+2
lim
n→∞
= lim
(n
+ 2) x
n +1
n→∞
n +1
n→∞
= x
= lim
n→∞
When x = 1, the series
∞
17.
∑ − (n + 1) diverges.
When x = −1, the series
∞
⎛
∑ (2n)!⎜
x⎞
⎟
⎝ 3⎠
n=0
n=0
Therefore, the interval of convergence is ( −1, 1).
lim
n=0
lim
n →∞
3x
= 0
+ 2)( 2n + 1)
n→∞
n
(2n + 2)!( x 3)
un + 1
= lim
→
∞
n
un
(2n)!( x 3)n
x5n
n!
∑
( 2n
Therefore, the interval of convergence is ( −∞, ∞).
∞
n +1
∑ (−1) (n + 1) diverges.
n=0
∞
(3 x )
(2n)!
un + 1
= lim
⋅
n
→
∞
un
(2n + 1)! (3x)n
n +1
lim
Interval: −1 < x < 1
15.
∞
∑ (2n)!
=
x5(n +1) ( n + 1)!
u n +1
x5
= lim
= lim
= 0
n
n →∞
n →∞ n + 1
un
5 n!
( 2n
+ 2)( 2n + 1) x
3
n +1
= ∞
The series converges only for x = 0.
The series converges for all x. The interval of
convergence is ( −∞, ∞).
18.
( −1)
∞
n
xn
∑ (n + 1)(n + 2)
n=0
(−1) x n +1 ⋅ (n + 1)(n + 2) = lim (n + 1) x = x
un + 1
= lim
n → ∞ ( n + 2)( n + 3)
n→∞
un
n +3
(−1)n x n
n +1
lim
n→∞
Interval: −1 < x < 1
When x = 1, the alternating series
When x = −1, the series
∞
( −1)
∑ (n + 1)(n
n=0
n
∞
+ 2)
converges.
∞
1
1
∑ (n + 1)(n + 2) converges by limit comparison to ∑ n2 .
n=0
n =1
Therefore, the interval of convergence is [−1, 1].
19.
∞
∑
(−1)
n +1 n
x
4n
n =1
Because the series is geometric, it converges only if x 4 < 1 or −4 < x < 4.
20.
∞
∑
(−1)n n!( x
n=0
− 5)
n
3n
(−1) (n + 1)!( x − 5) 3n +1 = lim ( n + 1)( x − 5) = ∞
u n +1
= lim
n
n
n →∞
n →∞
un
3
(−1) n!( x − 5) 3n
n +1
lim
n →∞
n +1
The series converges only for x = 5.
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
21.
∞
∑
(−1)n +1( x
− 4)
n
22.
n9 n
n =1
lim
n→∞
(x
− 3)
(−1)
un + 1
= lim
n
→
∞
un
n +1
∑ (n + 1)4n +1
( x − 4) ((n + 1)9n +1 )
(−1)n ( x − 4)n (n9n )
n+2
n +1
( x − 3)
un + 1
= lim
n +1
n
→
∞
un
( x − 3)
n→∞
= lim
(x
− 3)( n + 1)
4( n + 2)
n→∞
R = 9
R = 4
Interval: −5 < x < 13
Interval: −1 < x < 7
(−1)
∞
When x = 13, ∑
n +1 n
9
n9
n =1
=
n
(−1) (−9)
When x = −5, ∑
n
=
n9 n
n =1
∑
(−1)
n +1
When x = 7,
converges.
n
n =1
n +1
∞
∞
(−1)n +1 ( x
− 1)
4n +1
∞
∑ (n + 1)4n +1
=
n=0
∞
−1
diverges.
n =1 n
∑
When x = −1,
∞
(−4)
=
∞
x −3
4
1
∑ n + 1 diverges.
n=0
n +1
∑ (n + 1)4n +1
n=0
Therefore, the interval of convergence is ( −5, 13].
∞
⎡⎣( n + 2)4n + 2 ⎤⎦
⎡⎣( n + 1)4n + 1 ⎤⎦
n+2
lim
n ( x − 4)
1
x − 4
=
n +1 9
9
n→∞
∑
341
n=0
= lim
23.
∞
Power Series
=
∞
(−1)
n +1
∑ (n + 1) converges.
n=0
Therefore, the interval of convergence is [−1, 7).
n +1
n +1
n=0
un + 1
(−1) ( x − 1)
= lim
n→∞
un
n + 2
n+2
lim
n→∞
n+2
⋅
n +1
(−1) ( x
n +1
− 1)
n +1
= lim
n→∞
(n
+ 1)( x − 1)
n + 2
= x −1
R =1
Center: x = 1
Interval: −1 < x − 1 < 1 or 0 < x < 2
When x = 0, the series
∞
1
∑ n + 1 diverges by the integral test.
n=0
∞
∑
When x = 2, the alternating series
(−1)
n +1
n +1
n=0
converges.
Therefore, the interval of convergence is (0, 2].
24.
∞
∑
(−1)n +1 ( x
− 2)
n
n 2n
n =1
an +1
(−1) ( x − 2)
= lim
n →∞
an
(n + 1)2n +1
n+2
lim
n →∞
(−1) ( x − 2)
n +1
n +1
n2n
n
= lim
n →∞
x −2 n
x −2
=
2 n +1
2
x − 2
< 1 ⇒ −2 < x − 2 < 2 ⇒ 0 < x < 4
2
when x = 0,
∞
∑
n =1
(−1)n +1(−2)n
n2
n
=
∞
∑
(−1)(2n )
n2
n =1
n
=
∞
∑
(−1) diverges.
n =1
n
when x = 4,
∞
∑
n =1
(−1)n +1 2n
n2
n
=
∞
∑
n =1
(−1)n +1 converges.
n
Therefore the interval of convergence is (0, 4].
© 2010 Brooks/Cole, Cengage Learning
342
25.
Chapter 9
∞
⎛ x − 3⎞
⎟
3 ⎠
n =1
∑ ⎜⎝
Infinite Series
n −1
is geometric. It converges if
28.
∞
∑
(−1)n
n!
n=0
x −3
< 1 ⇒ x − 3 < 3 ⇒ 0 < x < 6.
3
x 2n
un + 1
(−1) x 2 n + 2 ⋅ n!
= lim
n
n→∞
un
(n + 1)!
(−1) x 2n
n +1
lim
n→∞
Therefore, the interval of convergence is (0, 6).
= lim
26.
∞
∑
n→∞
(−1)n x 2 n +1
Therefore, the interval of convergence is ( −∞, ∞).
2n + 1
n=0
u n +1
(−1) x 2 n + 3 ⋅ (2n + 1)
= lim
n →∞
un
(2n + 3) (−1)n x 2 n +1
n +1
lim
n →∞
(2n + 1) x 2
(2n + 3)
= lim
n →∞
29.
x 3n + 1
∞
∑ (3n + 1)!
n=0
= x2
lim
n→∞
x3n + 4 (3n + 4)!
un + 1
= lim 3n + 1
n
→
∞
un
x
(3n + 1)!
R =1
= lim
Interval: −1 < x < 1
When x = 1,
x2
= 0
n +1
n→∞
(−1)n
∞
x3
= 0
(3n + 4)(3n + 3)(3n + 2)
Therefore, the interval of convergence is ( −∞, ∞).
∑ 2n + 1 converges.
n=0
When x = −1,
∞
∑
n=0
(−1)
n +1
2n + 1
converges.
30.
∞
n
∑ n + 1(−2 x)
n!x n
n =1
Therefore, the interval of convergence is [−1, 1].
27.
∞
∑ (2n)!
lim
n→∞
n −1
= lim
n→∞
n =1
(n + 1)(−2 x) ⋅ n + 1
un + 1
= lim
n −1
n→∞
un
n + 2
n( −2 x)
n→∞
= lim
n→∞
R =
(−2 x)(n + 1)
n( n + 2)
2
= 2 x
31.
∞
∑
2⋅3⋅4
n =1
lim
n→∞
= 0
(n
+ 1) x n
n!
=
∞
∑ (n + 1) x n
n =1
(n + 2) x
an + 1
= lim
n → ∞ ( n + 1) x n
an
= lim
n→∞
n + 2
x = x
n +1
Converges if x < 1 ⇒ −1 < x < 1.
1
1
< x <
2
2
1
When x = − , the series
2
+ 1) x
n +1
1
2
Interval: −
(n
(2n + 2)(2n + 1)
Therefore, the interval of convergence is ( −∞, ∞).
n
lim
un + 1
(n + 1)!x n +1 ⋅ (2n)!
= lim
n→∞
un
(2n + 2)! n!x n
At x = ±1, diverges.
∞
n
∑ n + 1 diverges by the nth
Therefore the interval of convergence is ( −1, 1).
n =1
Term Test.
When x =
∞
∑
n =1
1
, the alternating series
2
(−1)n −1 n diverges.
n +1
⎛ 1 1⎞
Therefore, the interval of convergence is ⎜ − , ⎟.
⎝ 2 2⎠
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
32.
Power Series
343
( 2 n)
∞
2⋅4⋅6
lim
2 ⋅ 4 ( 2n)( 2n + 2) x 2 n + 3
3⋅5
un + 1
= lim
⋅
n
→
∞
un
3 ⋅ 5 ⋅ 7 ( 2n + 1)( 2n + 3) 2 ⋅ 4
∑ 3 ⋅ 5 ⋅ 7 (2n + 1) ( x 2n +1 )
n =1
n→∞
(2n + 1)
( 2 n) x 2 n + 1
= lim
n→∞
( 2 n + 2) x 2
(2n + 3)
= x2
R =1
When x = ±1, the series diverges by comparing it to
∞
1
∑ 2n + 1
n =1
which diverges.
Therefore, the interval of convergence is ( −1, 1).
33.
∞
∑
(−1)n +1 3 ⋅ 7 ⋅ 11 (4n
4
n =1
lim
n→∞
un + 1
( −1)
= lim
n
→
∞
un
n+2
( 4n
= lim
− 1)( x − 3)
n
n
⋅ 3 ⋅ 7 ⋅ 11
( 4n
− 1)( 4n + 3)( x − 3)
n +1
⋅
4n +1
+ 3)( x − 3)
( −1)
n +1
⋅ 3 ⋅ 7 ⋅ 11
( 4n
− 1)( x − 3)
n
= ∞
4
n→∞
4n
R = 0
Center: x = 3
Therefore, the series converges only for x = 3.
34.
n!( x + 1)
∞
n
∑ 1 ⋅ 3 ⋅ 5 (2n − 1)
n =1
an +1
(n + 1)!( x + 1)
= lim
n
→∞
1 ⋅ 3 ⋅ 5 ( 2n − 1)( 2n + 1)
an
n +1
lim
n →∞
(n)!( x
1⋅3⋅5
+ 1)
(n + 1)( x + 1) = 1 x + 1
= lim
n
→∞
2n + 1
2
(2n − 1)
n
1
x + 1 < 1 ⇒ −2 < x + 1 < 2 ⇒ −3 < x < 1.
2
n!2n
2 ⋅ 4 ⋅ 6 2n
At x = 1, an =
=
> 1,
1 ⋅ 3 ⋅ 5 ( 2n − 1)
1 ⋅ 3 ⋅ 5 ( 2n − 1)
Converges if
At x = −3,
an =
n!( −2)
n
( 2n
1⋅3
− 1)
= ( −1)
n
2 ⋅ 4 2n
,
1 ⋅ 3 ( 2n − 1)
diverges.
diverges.
Therefore, the interval of convergence is ( −3, 1).
35.
∞
∑
(x
n =1
− c)
c
n −1
n −1
( x − c) ⋅ c n −1
1
un + 1
= lim
=
x −c
n −1
n→∞
un
cn
c
−
x
c
(
)
n
lim
n→∞
R = c
Center: x = c
Interval: −c < x − c < c or 0 < x < 2c
When x = 0, the series
∞
∑ (−1)
n −1
diverges.
n =1
When x = 2c, the series
∞
∑ 1 diverges.
n =1
Therefore, the interval of convergence is (0, 2c).
© 2010 Brooks/Cole, Cengage Learning
344
36.
Chapter 9
(n!)
∞
k
xn
∑
n = 0 ( kn)!
Infinite Series
, k is a positive integer.
⎡( n + 1)!⎤⎦ x n +1
an +1
= lim ⎣
n →∞
an
⎣⎡k ( n + 1)⎤⎦!
k
lim
n →∞
Converges if
37.
∞
⎛ x⎞
∑⎜ ⎟
n=0 ⎝ k ⎠
x
(n!)k x n
(kn)!
= lim
n →∞
(k
(n + 1)
k
x
+ nk )( k − 1 + nk )
(1 +
nk )
=
x
kk
< 1 ⇒ R = kk.
kk
n
Because the series is geometric, it converges only if x k < 1 or − k < x < k .
Therefore, the interval of convergence is ( − k , k ).
38.
∞
∑
(−1)n +1( x
− c)
n
nc n
n =1
un + 1
(−1) ( x − c)
= lim
n
→
∞
un
(n + 1)c n +1
n+2
lim
n→∞
n +1
⋅
nc n
(−1) ( x
n +1
− c)
= lim
n
n→∞
n( x − c )
c( n + 1)
=
1
x −c
c
R = c
Center: x = c
Interval: −c < x − c < c or 0 < x < 2c
When x = 0, the p-series
∞
−1
diverges.
n =1 n
∑
When x = 2c, the alternating series
∞
∑
(−1)
n =1
n +1
n
converges.
Therefore, the interval of convergence is (0, 2c).
∞
39.
∑
k ( k + 1)
n→∞
+ n − 1) x n
n!
n =1
lim
(k
k ( k + 1)
un + 1
= lim
n→∞
un
(k
+ n − 1)( k + n) x n + 1
(n
+ 1)!
⋅
k ( k + 1)
n!
( k + n) x = x
= lim
n
n→∞
n +1
( k + n − 1) x
R =1
When x = ±1, the series diverges and the interval of convergence is ( −1, 1).
⎡ k ( k + 1) ( k + n − 1)
⎤
≥ 1⎥
⎢
1
⋅
2
n
⎣
⎦
40.
n!( x − c)
∞
n
∑ 1 ⋅ 3 ⋅ 5 (2n − 1)
n =1
1 ⋅ 3 ⋅ 5 ( 2n − 1)
(n + 1)!( x − c)
(n + 1)( x − c) = 1 x − c
un + 1
= lim
⋅
= lim
n
n
→
∞
→
∞
un
n!( x − c)
1 ⋅ 3 ⋅ 5 ( 2n − 1)( 2n + 1)
2n + 1
2
n +1
lim
n→∞
R = 2
Interval: −2 < x − c < 2 or c − 2 < x < c + 2
The series diverges at the endpoints. Therefore, the interval of convergence is (c − 2, c + 2).
⎡ n!(c + 2 − c)n
⎤
2 ⋅ 4 ⋅ 6 ( 2n)
n!22
⎢
=
=
> 1⎥
1 ⋅ 3 ⋅ 5 ( 2n − 1)
1 ⋅ 3 ⋅ 5 ( 2n − 1)
⎢⎣1 ⋅ 3 ⋅ 5 ( 2n − 1)
⎥⎦
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
41.
∞
∑
n=0
∞
42.
∑
xn
x
x2
=1+ +
+
1
2
n!
(−1)
n +1
(n + 1) x n =
n=0
43.
∞
∑
n=0
x n −1
∞
∑ (n − 1)!
=
48. (a) f ( x) =
n =1
∞
(b) f ′( x) =
∑ (−1) (n) x n −1
n
∞
∑
(−1)
n
x 2 n +1
2n + 1
n=0
(−1)
∞
x
49. g (1) =
⎛ x⎞
∑ ⎜⎝ 3 ⎟⎠ , (−3, 3)
(Geometric)
n=0
(b) f ′( x) =
∞
n⎛ x ⎞
∑ 3 ⎜⎝ 3 ⎟⎠
∑
(c) f ′′( x) =
n=2
n −1
(d)
∫ f ( x) dx
n=0
n +1
⎡
3 ⎛ −3 ⎞
⎢∑
=
⎜ ⎟
⎢⎣ n + 1⎝ 3 ⎠
46. (a) f ( x) =
∞
∑
n5
5
n =1
(c) f ′′( x) =
∞
∑
∫
47. (a) f ( x) =
∑
(−1) (n
n −1
n−2
, (0, 10)
− 1)
n +1
, (0, 2]
∞
n +1
n
∑ (−1) ( x − 1) , (0, 2)
∑ (−1)
n( x − 1)
n −1
, (0, 2)
n =1
(d)
∫ f ( x) dx =
(−1) ( x − 1)
∑ (n + 1)(n + 2) , [0, 2]
n =1
∞
n
4
9
+
diverges. Matches (b).
∞
∑ (− 23 )
n
alternating. Matches (d).
53. g
∞
( 18 ) = ∑ ⎣⎡2( 18 )⎦⎤
n
=
n=0
∞
∑ ( 14 )
n
=1+
n=0
1
4
+
1
16
+ ⋅ ⋅ ⋅,
converges
∞
( ) ∑ ⎡⎣2(− 18 )⎤⎦
n=0
(c) f ′′( x) =
∞
54. g − 18 =
, (0, 10)
− 1)( x − 5)
n +1
n +1
+
S1 = 1, S 2 = 1.25, S3 = 1.3125 Matches (b).
, (0, 10]
(−1)n +1 ( x − 5)n +1 , 0, 10
∑ n(n + 1)5n [ ]
n =1
(−1)n +1 ( x
2
3
=1+
n=0
⎥⎦
5n
∞
+
n
∞
∑ (− 14 )
=
n=0
n=0
n
=1−
1
4
+
1
16
−
,
converges
n +1
n=0
(b) f ′( x) =
, [−3, 3)
∞
f ( x) dx =
∞
n
n
n=2
(d)
n
∑ ( 3.13)
52. g ( −2) =
n +1
− 5)
n +1
∑
1
9
, [1, 3]
n=0
(−1)n +1 3 , converges⎥⎤
n
(−1) ( x
∞
∑ ( 23 )
50. g ( 2) =
51. g (3.1) =
, ( −3, 3)
n +1
− 5)
n +1
n =1
(b) f ′( x) =
∑
(−1) ( x
∞
n−2
3 ⎛ x⎞
⎜ ⎟
n + 1⎝ 3 ⎠
∑
=
+
n +1
S1 = 1, S 2 = 1.67. Matches (a).
, ( −3, 3)
n( n − 1) ⎛ x ⎞
⎜ ⎟
9
⎝ 3⎠
∞
1
3
=1+
n=0
n =1
∞
n
n +1
S1 = 1, S2 = 1.33. Matches (c).
n
∞
∞
∑ ( 13 )
( −1) ( x − 2)
n( n + 1)
n =1
∞
∑
n=0
Replace n with n − 1.
45. (a) f ( x) =
∞
n +1
n−2
∑ (−1) (n − 1)( x − 2) , (1, 3)
∫ f ( x) dx =
n −1 2 n −1
2n − 1
n =1
∞
n=2
(d)
∑
, (1, 3]
n +1
n −1
∑ (−1) ( x − 2) , (1, 3)
(c) f ′′( x) =
x 2 n −1
∑ (2n − 1)!
n =1
∞
=
n
345
n =1
Replace n with n − 1.
44.
− 2)
n
n =1
n =1
x 2 n +1
=
(2n + 1)!
(−1)n +1 ( x
∞
∑
Power Series
n +1
n+2
S1 = 1, S 2 = 0.75, S3 = 0.8125 Matches (c).
55. g
∞
(169 ) = ∑ ⎡⎣2(169 )⎤⎦
n
=
n=0
S1 = 1, S 2 =
56. g
17
8
∞
∞
∑ ( 89 ) ,
n
diverges
n=0
Matches (d).
( 83 ) = ∑ ⎡⎣2( 83 )⎤⎦
n
=
n=0
∞
∑ ( 43 )
n
,
converges
n=0
S1 = 1, S2 = 1.75 Matches (a).
57. A series of the form
∞
∑ an ( x − c)
n
= a0 + a1 ( x − c) + a2 ( x − c) +
2
n=0
+ an ( x − c) +
n
is called a power series centered at c, where c is constant.
© 2010 Brooks/Cole, Cengage Learning
346
Chapter 9
Infinite Series
58. The set of all values of x for which the power series
converges is the interval of convergence. If the power
series converges for all x, then the radius of convergence
is R = ∞. If the power series converges at only c, then
R = 0. Otherwise, according to Theorem 8.20, there
exists a real number R > 0 (radius of convergence) such
that the series converges absolutely for x − c < R and
63. (a) f ( x) =
n=0
lim
n→∞
n→∞
g ( x) =
60. You differentiate and integrate the power series term by
term. The radius of convergence remains the same.
However, the interval of convergence might change.
n→∞
∞
∑
n =1
n→∞
x
⎛ x⎞
∑ ⎜⎝ 2 ⎟⎠ Geometric: 2 < 1 ⇒ x < 2
n =1
(b)
∑
∞
(−1)n x n converges for
n
n =1
∞
(c)
∑ (2 x + 1)
n
−1 < x ≤ 1
Geometric:
n =1
(d)
∑
n =1
(x
− 2)
n4n
∞
(−1)n x 2 n −1
(c) g ′( x) =
∑ (2n − 1)!
n =1
= g ( x)
=
(−1)n x 2 n +1
n = 0 ( 2 n + 1)!
∞
= −∑
∞
(−1)n +1 x 2n +1
∑ (2n + 1)!
n=0
= − f ( x)
(d) f ( x) = sin x and
g ( x) = cos x
64. (a) f ( x) =
∞
∑
n=0
2 x + 1 < 1 ⇒ −1 < x < 0
∞
(−1)n x 2 n
∑ (2n)!
n=0
n
∞
∞
(b) f ′( x) =
62. Many answers possible.
(a)
1
= 0
2n + 2
Therefore, the interval of convergence is ( − ∞ ∞).
1
convergence is absolute.
x 2n
= lim
∑ n diverges.
xn
converges for −1 ≤ x ≤ 1. At x = ±1, the
n2
n
(−1) x 2 n + 2 ⋅ (2n)!
un + 1
= lim
n→∞
un
(2n + 2)! (−1)n x 2 n
xn
converges for −1 ≤ x < 1. At x = −1, the
n
convergence is conditional because
(−1)
∞
∑ (2n)!
n=0
n +1
lim
61. Answers will vary.
n =1
x2
= 0
(2n + 2)(2n + 3)
Therefore, the interval of convergence is ( − ∞ ∞).
59. A single point, an interval, or the entire real line.
∞
(2n + 1)!
un + 1
x2n + 3
= lim
⋅
n → ∞ ( 2 n + 3)!
un
x 2 n +1
= lim
diverges for x − c > R.
∑
x 2 n +1
∞
∑ (2n + 1)!
lim
n
n→∞
converges for −2 ≤ x < 6
xn
n!
un + 1
x n +1
n!
= lim
⋅ n
n
→
∞
un
(n + 1)! x
= lim
n→∞
x
= 0
n +1
The series converges for all x. Therefore, the interval
of convergence is ( − ∞, ∞).
∞
∑
(b) f ′( x) =
n =1
(c)
f ( x) =
∞
∑
n =1
nx n −1
=
n!
x n −1
∞
∑ (n − 1)!
n =1
∞
∑
n=0
xn
= f ( x)
n!
x
x
x
x4
= 1+ x +
+
+
+
n!
2! 3! 4!
n
2
=
3
f (0) = 1
(d) f ( x) = e x
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
65. y =
y′ =
y′′ =
(−1)n x 2 n +1
∞
∑
n = 0 ( 2n + 1)!
(−1) (2n) x 2 n −1
∑
(2n)!
n =1
=
(−1) x 2n −1
∑ (2n − 1)!
n =1
+
(−1)
∑ ( 2n
n =1
n
y′ =
y′′ =
n
(−1)n x 2 n
∑ (2n)!
n=0
∞
x 2 n −1
− 1)!
=
(−1)n (2n − 1) x 2n − 2
(2n − 1)!
n =1
∞
∞
(−1)n x 2 n − 2
∑
n =1 ( 2 n − 2)!
x2n
y′ =
n=0
=
∞
∑
∞
∑
=
2 n!
∞
2n( 2n − 1) x 2 n − 2
∑
y′ =
y′′ =
−
∞
∑
∞
∑
2
⎣
n +1
(n
+ 1)!
n=0
=
x 2 n −1
∞
∑ (2n − 1)!
= y
n =1
x2n
( 2n
=
x 2n − 2
∞
∑ (2n − 2)!
n =1
− 2) x 2 n −1
∑ (2n − 2)!
n =1
∞
(2n − 1) x 2 n − 2
∑ (2n − 1)!
n =1
=
x 2 n −1
∞
∑ (2n − 1)!
n =1
=
x2n − 2
∞
∑ (2n − 2)!
= y
n =1
2n( 2n − 1) x 2 n − 2
∞
2n n!
2nx 2 n
−
2n n!
( 2n
∞
x 2n
∑ 2n n!
n=0
+ 1) x 2 n
2n n!
n=0
⎡ ( 2n + 2)( 2n + 1) x 2 n
∞
−
( 2n
+ 1) x 2 n
n
2 n!
⋅
2( n + 1) ⎤
⎥
2( n + 1) ⎥⎦
2( n + 1) x ⎡⎣( 2n + 1) − ( 2n + 1)⎤⎦
= 0
2n + 1 ( n + 1)!
2n
(−1)n
∞
x4n
∑ 22n n! ⋅ 3 ⋅ 7 ⋅ 11 (4n − 1)
n =1
(−1)n 4nx 4 n −1
∑ 22n n! ⋅ 3 ⋅ 7 ⋅ 11 (4n
n =1
∞
∞
(−1)
n
− 1)
4n ( 4n − 1) x 4 n − 2
∑ 22n n! ⋅ 3 ⋅ 7 ⋅ 11 (4n − 1)
n =1
y′′ + x 2 y = − x 2 +
∞
∑
n=2
=
n =1
2n n!
∑ ⎢⎢
n=0
y =1+
∞
∑
x2n
= 0
n =1
−
∞
∑ (2n)!
=
y′′ − y = 0
∑
y′′ =
n =1
n=0
=
2nx 2 n −1
2n n!
+ 1) x
2n
∑ (2n)!
n =1
∞
y′′ =
∑ (2n − 2)!
n =1
n
( 2n ) x 2 n − 1
∞
y′ =
(−1)n −1 x 2 n − 2
∞
∞
n=0
(−1)n x 2 n − 2
∞
( 2n
∑ (2n)!
68. y =
∑ (2n − 2)!
n =1
=
2n( 2n − 1) x 2 n − 2
n =1
70.
+
n =1
y′′ − xy′ − y =
= 0
− 1)!
(−1)n x 2 n −1
∑ (2n − 1)!
n =1
∑
∑ 2n n!
x
∞
=
n =1
∞
y′′ − y = 0
n −1 2 n −1
∞
(−1)n −1 x 2n − 2
∑ (2n − 2)!
n =1
(−1)n (2n) x 2n −1
∑
(2n)!
n =1
∞
y′′ =
∞
x 2 n −1
∞
∑
n = 0 ( 2n + 1)!
y′ =
∞
∞
y′′ + y =
69. y =
(−1)
∑ ( 2n
n =1
∞
(−1)n x 2 n
∑ (2n)!
n=0
∞
347
∑ (2n − 1)!
=
n=0
=
∞
x 2 n +1
∞
∑ (2n + 1)!
67. y =
(−1)n (2n + 1) x 2 n
∑ (2n + 1)!
n=0
∞
y′′ + y =
66. y =
(−1)n −1 x 2 n −1
∞
∑ (2n − 1)!
n =1
=
Power Series
∞
(−1)
n
4nx 4 n − 2
22 n n! ⋅ 3 ⋅ 7 ⋅ 11
(−1)n +1 4(n
= − x2 +
( 4n
+ 1) x 4 n + 2
− 5)
+
∞
(−1)
n
4nx 4 n − 2
∑ 2n
(4n − 5)
n = 2 2 n! ⋅ 3 ⋅ 7 ⋅ 11
∞
(−1)
n
x 4n + 2
∑ 22n n! ⋅ 3 ⋅ 7 ⋅ 11 (4n − 1)
n =1
∞
(−1)n +1 x 4 n + 2
+ x2
22 ( n + 1)
∑ 22n + 2 (n + 1)! ⋅ 3 ⋅ 7 ⋅ 11 (4n − 1) − ∑ 22n n! ⋅ 3 ⋅ 7 ⋅ 11 (4n − 1) 22 (n + 1)
n =1
n =1
= 0
© 2010 Brooks/Cole, Cengage Learning
348
Chapter 9
71. J 0 ( x) =
Infinite Series
(−1)k x 2 k
2
2k
k = 0 2 ( k!)
∞
∑
(−1) x 2k + 2 ⋅ 22 k (k!) = lim (−1) x 2 = 0
uk + 1
= lim
2
k
2
k →∞
uk
(−1) x 2 k k → ∞ 22 (k + 1)
22 k + 2 ⎣⎡( k + 1)!⎤⎦
k +1
(a) lim
k →∞
2
Therefore, the interval of convergence is −∞ < x < ∞.
J0 =
(b)
∞
∑ (−1)
k
4k ( k!)
k =0
J 0′ =
∞
∑ (−1)
k
∞
∑ (−1)
k
∞
∑ (−1)
∞
∑
( 2k
k +1 2
4
(c) P6 ( x) = 1 −
k +1
∞
∑ (−1)
=
k +1
( 2k
+ 1)!k!
+
∞
∑ (−1)
2
+ 2)( 2k + 1) x 2 k
4k + 1 ⎡⎣( k + 1)!⎤⎦
k =0
+ 1) x 2 k + 2
(k
+ 2) x 2 k + 1
4k + 1 ⎡⎣( k + 1)!⎤⎦
k =0
2
( 2k
k +1
k +1
k =0
2 x2k + 2
+
4 ( k + 1)!k!
k +1
2
∞
∑ (−1)
k =0
k
x 2k + 2
4k ( k!)
2
⎤
x 2k + 2 ⎡
2( 2k + 1)
2
+ ( −1)
+ 1⎥
⎢( −1)
2
4( k + 1)
4( k + 1)
4 ( k!) ⎢⎣
⎥⎦
(−1)
k
k
k =0
=
∞
∑ (−1)
=
4k ( k!)
k =0
=
2
2k ( 2k − 1) x 2 k − 2
k =1
x 2 J 0′′ + xJ 0′ + x 2 J 0 =
2
2kx 2 k −1
4k ( k!)
k =1
J 0′′ =
x2k
(−1)k x 2k + 2 ⎡ −4k − 2
2
⎢
4k ( k!) ⎣ 4k + 4
k =0
∞
∑
−
2
4k + 4 ⎤
+
= 0
4k + 4
4k + 4 ⎥⎦
x2
x4
x6
+
−
4
64
2304
3
−6
6
−5
(d)
1
∫ 0 J 0dx
=
(−1)k x 2 k
∫ 0 k∑= 0 4k (k!)2
1 ∞
1
⎡∞
(−1)k x 2k +1 ⎥⎤ =
dx = ⎢ ∑
2
k
⎣⎢k = 0 4 ( k!) ( 2k + 1) ⎥⎦ 0
∞
∑ 4k
k =0
(−1)k
( k!)2 (2k
+ 1)
= 1−
1
1
+
≈ 0.92
12 320
(integral is approximately 0.9197304101)
(−1) x 2k
2k +1
k!( k + 1)!
k =0 2
∞
72. J1 ( x) = x ∑
k
=
∞
(−1)
k
x 2k +1
∑ 22k +1 k!(k
k =0
+ 1)!
22 k + 1 k!( k + 1)!
(−1) x 2k + 3
(−1) x 2
uk + 1
= lim 2 k + 3
⋅
= lim 2
= 0
k
k →∞ 2
k → ∞ 2 ( k + 2)( k + 1)
uk
(k + 1)!( k + 2)! (−1) x 2 k +1
k +1
(a) lim
k →∞
Therefore, the interval of convergence is −∞ < x < ∞.
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
J1 ( x ) =
(b)
+ 1!)
J1′ ( x) =
(−1) (2k + 1) x 2k
2k +1
k!( k + 1)!
k =0 2
J1′′ ( x) =
∑
k
∞
∑
(−1)k (2k
∞
2
k =1
+ 1)( 2k ) x 2 k −1
2k +1
k!( k + 1)!
x 2 J1′′ + xJ1′ + ( x 2 − 1) J1 =
∞
∑
(−1) (2k
k
2
k =1
+
2 k +1
+ 1)( 2k ) x 2 k + 1
(−1)
∞
k
∑
(−1) (2k
+ 1)!
−
2
(−1)
∞
(−1)
k
∞
∑
(−1)
k
(−1)
k
k =1
=
∞
∑
=
∑
(−1)k (2k
22 k + 1 k!( k + 1)!
k =1
x 2 k + 1 ⎡⎣( 2k + 1)( 2k ) + ( 2k + 1) − 1⎤⎦
+
22 k + 1 k!( k + 1)!
x
2 k +1
+ 1) x 2 k + 1
−
x
−
2
(−1)k x 2k +1 ⎥⎤
∑ 2k +1 k!(k + 1)!⎥
k =1 2
⎦
∞
4k ( k + 1)
∞
+
(−1)
k
x
(−1)
∞
k
x 2k + 3
∑ 22k +1 k!(k
k =0
+ 1)!
2k + 3
∑ 22k +1 k!(k
k =0
+ 1)!
∞
(−1)k +1 x 2 k + 3 + ∞ (−1)k x 2 k + 3
∑
+ 1)! k = 0 22 k + 1 k!( k + 1)!
∑ 22k +1 k!(k
∞
∑
(−1)
x 2 k + 3 ⎡⎣( −1) + 1⎤⎦
= 0
22 k + 1 k!( k + 1)!
n=0
J 0′ ( x) =
∞
(−1)k x 2k +1 + ∞ (−1)k x 2 k + 3
∑
− 1)!k! k = 0 22 k + 1 k!( k + 1)!
∞
k =0
=
+ 1)!
∑ 22 k −1 ( k
k =1
=
x 2k +1
+ 1)!
22 k + 1 k!( k + 1)!
k =1
k!( k + 1)!
x2k + 3
∑ 22k +1 k!(k
k =0
=
k
k =0
⎡ ∞ ( −1) ( 2k + 1)( 2k ) x 2 k + 1
x
= ⎢∑
+ +
22 k + 1 k!( k + 1)!
2
⎢⎣k =1
∞
2k +1
∑ 22k +1 k!(k
k
+
+ 1) x 2 k + 1
k
k =0
x2k + 3
∑ 22k +1 k!(k
∞
+
k!( k + 1)!
k =0
(d)
349
(−1)k x 2 k +1
∞
∑ 22k +1 k!(k
k =0
(c) P7 ( x) =
Power Series
k
x
1 3
1 5
1
−
x +
x −
x7
2 16
384
18,432
(−1)k +1 2(k + 1) x 2 k +1
2k + 2
(k + 1)!(k + 1)!
k =0 2
∞
∑
(−1)k x 2k +1
k!( k + 1)!
k =0 2
∞
− J1 ( x ) = − ∑
2 k +1
=
=
∞
∞
(−1)k +1 x 2 k +1
∑ 22k +1 k!(k
k =0
(−1)k +1 x 2 k +1
∑ 22k +1 k!(k
k =0
+ 1)!
+ 1)!
Note: J 0′ ( x) = − J1 ( x)
4
−6
6
−4
73. f ( x) =
∞
x2n
∑ (−1) (2n)!
n=0
n
= cos x
74. f ( x) =
2
− 2␲
n
x 2 n +1
= sin x
2
2␲
−2
∞
∑ (−1) (2n + 1)!
n=0
−␲
␲
−2
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
350
∞
Infinite Series
∑ (−1)
75. f ( x) =
n
xn =
n=0
∞
n
Geometric
78.
n=0
1
=
∞
∑ ( − x)
1 − ( − x)
⎛ 1 1⎞
converges on ⎜ − , ⎟.
⎝ 3 3⎠
n
n=0
1
for −1 < x < 1
1+ x
=
∑ (3 x )
n
1 ∞ ⎛ ⎛ 1 ⎞⎞
: ∑ ⎜ 3⎜ ⎟ ⎟ =
6 n =0 ⎝ ⎝ 6 ⎠⎠
(a) x =
n
∞
1
⎛1⎞
∑ ⎜⎝ 2 ⎟⎠ = 1 − (1 2) = 2
n=0
3
3
−1
1
−1
0
6
0
1
(b) x = − :
6
x 2 n +1
76. f ( x) = ∑ ( −1)
= arctan x, −1 ≤ x ≤ 1
2n + 1
n=0
∞
n
␲
2
2.5
n
∞
⎛ x⎞
77. ∑ ⎜ ⎟ ,
n=0 ⎝ 4 ⎠
n
=
n=0
−1
∞
⎛ 1⎞
∑ ⎜⎝ − 2 ⎟⎠
n
=
n=0
1
2
=
1 + (1 2) 3
6
− 0.5
␲
2
(c) The alternating series converges more rapidly. The
partial sums in (a) approach the sum 2 from below.
The partial sums in (b) alternate sides of the
2
horizontal line y = .
3
(− 4, 4)
⎛ ( 5 2) ⎞
∑⎜ 4 ⎟ =
n=0 ⎝
⎠
n
∞
(a)
⎛ ⎛ 1 ⎞⎞
1.5
−2.5
−
∞
∑ ⎜⎝ 3⎜⎝ − 6 ⎟⎠ ⎟⎠
n
∞
1
8
⎛5⎞
∑⎜ ⎟ = 1−58 = 3
n=0 ⎝ 8 ⎠
N
4
⎛
2⎞
∑ ⎜⎝ 3 ⋅ 3 ⎟⎠
(d)
n
=
n=0
0
6
N
∑ 2n
> M
n=0
M
10
100
1000
10,000
N
3
6
9
13
0
⎛ ( −5 2) ⎞
⎟ =
4 ⎠
n=0 ⎝
n
∞
∑⎜
(b)
∞
⎛ 5⎞
∑ ⎜⎝ − 8 ⎟⎠
n
=
n=0
1
8
=
1+58
13
79. False;
∞
∑
(−1)
xn
n2n
n =1
1
n
converges for x = 2 but diverges for x = −2.
80. False; it is not possible. See Theorem 9.20.
0
6
0
(c) The alternating series converges more rapidly. The
partial sums of the series of positive terms
approaches the sum from below. The partial sums of
the alternating series alternate sides of the horizontal
line representing the sum.
(d)
83. lim
n→∞
81. True; the radius of convergence is R = 1 for both series.
82. True
1
∫0
f ( x) dx =
1⎛ ∞
⎞
n
∫ 0 ⎜⎝ n∑= 0 an x ⎟⎠ dx
1
M
10
100
1000
10,000
N
5
14
24
35
an + 1
(n + 1 + p)! x n +1
= lim
n
→
∞
an
(n + 1)!( n + 1 + q)!
⎡ ∞ a x n +1 ⎤
= ⎢∑ n
⎥ =
⎣n = 0 n + 1 ⎦ 0
(n + p)! x n
n!( n + q)!
= lim
n→∞
(n
(n
+ 1 + p) x
+ 1)( n + 1 + q)
∞
a
∑ n +n 1
n=0
= 0
So, the series converges for all x : R = ∞.
© 2010 Brooks/Cole, Cengage Learning
Section 9.8
Power Series
351
84. (a) g ( x) = 1 + 2 x + x 2 + 2 x3 + x 4 +
S2 n = 1 + 2 x + x 2 + 2 x3 + x 4 +
∞
lim S2 n =
n→∞
∑
n=0
+ x 2 n + 2 x 2 n +1 = (1 + x 2 + x 4 +
+ x 2 n ) + 2 x(1 + x 2 + x 4 +
+ x2n )
∞
x2n + 2 x ∑ x2n
n=0
Each series is geometric, R = 1, and the interval of convergence is ( −1, 1).
(b) For x < 1, g ( x) =
85. (a) f ( x) =
1
1
1 + 2x
.
+ 2x
=
2
2
1− x
1− x
1 − x2
∞
∑ cn x n , cn + 3
= cn
n=0
= c0 + c1x + c2 x 2 + c0 x3 + c1 x 4 + c2 x5 + c0 x 6 +
S3n = c0 (1 + x3 +
+ x3n ) + c1x(1 + x3 +
∞
∞
∞
n=0
n=0
n=0
+ x3n ) + c2 x 2 (1 + x3 +
+ x 3n )
lim S3n = c0 ∑ x3n + c1 x ∑ x3n + c2 x 2 ∑ x3n
n→∞
Each series is geometric, R = 1, and the interval of convergence is ( −1, 1).
(b) For x < 1, f ( x) = c0
86. For the series
lim
n→∞
n→∞
∑ cn x n ,
an + 1
c x n +1
c
cn
= lim n + 1 n
= lim n + 1 x < 1 ⇒ x <
= R
n→∞
n→∞ c
an
cn x
c
n
n +1
For the series
lim
c + c1 x + c2 x 2
1
1
1
+ c1 x
+ c2 x 2
= 0
.
3
3
3
1− x
1− x
1− x
1 − x3
∑cn x 2n ,
bn + 1
c x 2n + 2
c
cn
= lim n + 1 2 n
= lim n + 1 x 2 < 1 ⇒ x 2 <
= R ⇒ x <
n→∞
n→∞ c
bn
cn x
c
n
n +1
87. At x = x0 + R,
∞
n
∑ cn ( x − x0 )
=
n=0
At x = x0 − R,
∞
n
∑ cn ( x − x0 )
n=0
R.
∞
∑ cn R n , diverges.
n=0
=
∞
∑ cn (− R)
n
, converges.
n=0
Furthermore, at x = x0 − R,
∞
n
∑ cn ( x − x0 )
n=0
=
∞
∑ Cn R n , diverges.
n=0
So, the series converges conditionally at x0 − R.
© 2010 Brooks/Cole, Cengage Learning
352
Chapter 9
Infinite Series
Section 9.9 Representation of Functions by Power Series
1. (a)
1
14
=
4− x
1 − ( x 4)
=
a
=
1− r
3. (a)
∞
⎛ 1 ⎞⎛ x ⎞
∑ ⎜⎝ 4 ⎟⎜
⎟
⎠⎝ 4 ⎠
n
=
n=0
∞
xn
∑ 4 n +1
=
n=0
∑
n=0
n
1⎛ x ⎞
⎜− ⎟ =
2⎝ 2 ⎠
n
∞
∑
=
3( −1) x n
n
4
n=0
n +1
3 3x
3x 2
−
+
−
4 16
64
(b) 4 + x 3
3
3+ x
4
3
− x
4
3
3x 2
− x −
4
16
3x 2
16
3x 2
3 x3
+
16
64
1
12
a
=
=
2+ x
1 − ( − x 2)
1− r
∞
3⎛ x ⎞
This series converges on ( −4, 4).
x
x2
1
+
+
+
4 16
64
(b) 4 − x 1
x
1−
4
x
4
x
x2
−
4 16
x2
16
x2
x3
−
16
64
=
∞
∑ 4 ⎜⎝ − 4 ⎟⎠
n=0
This series converges on ( −4, 4).
2. (a)
3
34
a
=
=
4+ x
1 − ( − x 4)
1− r
∞
∑
n=0
4. (a)
2
25
a
=
=
5− x
1 − ( x 5)
1− r
(−1)n x n
2
=
n +1
∑
n=0
This series converges on ( −2, 2).
1
x
x2
x3
− +
−
+
2
4
8
16
(b) 2 + x 1
x
1+
2
x
−
2
x
x2
− −
2
4
x2
4
x2
x3
+
4
8
x3
−
8
x3
x4
−
−
8
16
∞
n
2⎛ x ⎞
⎜ ⎟ =
5⎝ 5 ⎠
∞
2 xn
∑ 5n + 1
n=0
This series converges on ( −5, 5).
2
2x
2x2
+
+
+
5
25 125
(b) 5 − x 2
2
2− x
5
2
x
5
2x
2 x2
−
5
25
2x2
25
5.
1
1
=
=
3− x
2 − ( x − 1)
=
∞
∑
n=0
n
12
a
=
1− r
⎛ x − 1⎞
1−⎜
⎟
⎝ 2 ⎠
1 ⎛ x − 1⎞
⎜
⎟ =
2⎝ 2 ⎠
Interval of convergence:
∞
∑
n=0
(x
− 1)
n
2n +1
x −1
< 1 ⇒ ( −1, 3)
2
© 2010 Brooks/Cole, Cengage Learning
Section 9.9
6.
4
4
=
=
5− x
8 − ( x + 3)
∞
=
n
∞
1⎛ x + 3⎞
⎟ =
8 ⎠
∑ 2 ⎜⎝
n=0
Interval of convergence:
7.
12
⎛x +
1−⎜
⎝ 8
∑
n=0
+ 3)
=
12.
=
23 n + 1
∞
13.
∞
n
∑ (5 x )
=
5
5
−5 9
a
=
=
=
2
2x − 3
−9 + 2( x + 3)
1
−
r
1 − ( x + 3)
9
∞
n=0
14.
10.
2
15 3
( x + 3) < 1 ⇒ ⎛⎜ − , ⎞⎟
9
⎝ 2 2⎠
∞
⎡ 2
⎤
∑ ⎢⎣− 3 ( x − 2)⎥⎦
n
∞
∑
( −2) ( x
n
− 2)
15.
3n
n=0
Interval of convergence: x − 2 <
=
2 ∞ ⎛ 2x ⎞
∑ ⎜− ⎟
3n=0 ⎝ 3 ⎠
=
∑
∞
n=0
( −1)
n
n =0
n=0
⎡ 1
⎣(
)
⎤
− 1⎥ x n
⎥⎦
n
∞
=
∑ ⎢⎜⎝ − 2 ⎟⎠
∞
⎡⎛ 1 ⎞ n
⎢⎣
⎤
+ 3n + 1 ⎥ x n
⎥⎦
x
< 1 and
2
1⎞
⎟
3⎠
2
1
1
=
+
1 − x2
1− x 1+ x
3
⎛1 7⎞
⇒ ⎜ , ⎟
2
⎝ 2 2⎠
2
23
23
a
11.
=
=
=
2
2x ⎞
2x + 3
1
r
⎛
−
1+ x
1 − ⎜− ⎟
3
3
⎝
⎠
∞
∞
n
⎛ x⎞
∑ ⎜⎝ − 2 ⎟⎠ + 3 ∑ (3x)
n=0
n=0
⎛ 1
3x < 1 ⇒ ⎜ − ,
⎝ 3
n
∞
∑ x n = ∑ ⎢⎢ −3 n
x
< 1 and x < 1 ⇒ ( −1, 1)
3
Interval of convergence:
n=0
=
−
=
n=0
3
3
1
a
=
=
=
2x − 1
3 + 2( x − 2)
1 + ( 2 3)( x − 2)
1− r
=
n
3x − 8
2
3
=
−
3x 2 + 5 x − 2
x + 2 3x − 1
1
3
=
+
1 − ( − x 2) 1 − 3 x
2n
n
( x + 3)
9n + 1
Interval of convergence:
⎛ x⎞
Interval of convergence:
5 ∞ ⎛2
2
⎞
∑ ⎜ ( x + 3) ⎟⎠ , 9 ( x + 3) < 1
9n=0 ⎝ 9
= −5 ∑
∞
∑ ⎜⎝ − 3 ⎟⎠
n =0
n
= −
3
20
( x − 3) < 1 ⇒ ⎛⎜ 3, ⎞⎟
11
⎝ 3⎠
4x
3
1
=
+
x2 + 2 x − 3
x + 3 x −1
1
−1
=
+
1 − ( − x 3) 1 − x
n=0
⎛1 1⎞
Interval of convergence: 5 x < 1 ⇒ ⎜ , ⎟
⎝5 5⎠
9.
11n +1
n=0
n
1⎞
⎟
3⎠
n
(−3)n ( x − 3)n
Interval of convergence: −
∑ (3 x )
1
a
=
=
1 − 5x
1− r
4 ∞ ⎡−3( x − 3) ⎤
∑⎢
⎥
11 n = 0 ⎣ 11
⎦
= 4∑
n=0
353
4
4
4 11
a
=
=
=
3x + 2 11 + 3( x − 3) 1 − (−3 11)( x − 3) 1 − r
n
⎛1
Interval of convergence: 3 x < 1 ⇒ ⎜ ,
⎝3
8.
a
1− r
x +3
< 1 ⇒ ( −11, 5)
8
∞
1
a
=
=
1 − 3x
1− r
(x
3⎞
⎟
⎠
Representation of Functions by Power Series
=
∑ (1 + (−1)
∞
n=0
n
)x
n
∞
= 2 ∑ x2n
n=0
Interval of convergence: x 2 < 1 or (1, 1) because
lim
n→∞
un + 1
2 x2n + 2
= lim
= x2
n → ∞ 2 x2n
un
n
2n +1
3n + 1
Interval of convergence:
xn
−2 x
⎛ 3 3⎞
< 1 ⇒ ⎜− , ⎟
3
⎝ 2 2⎠
© 2010 Brooks/Cole, Cengage Learning
354
16.
Chapter 9
5
=
5 + x2
Infinite Series
1
a
=
=
2
⎛ −x ⎞
1− r
1−⎜
⎟
⎝ 5 ⎠
Interval of convergence:
17.
1
=
1+ x
1
=
1− x
∞
∑ (−1)
n
⎛ x2 ⎞
∞
∑ ⎜− 5 ⎟
n=0
⎝
n
=
⎠
∞
⎛ −1 ⎞
∑ ⎜⎝ 5 ⎟⎠
n
x2n
n=0
x2
< 1 ⇒ −5 < x 2 < 5 ⇒ − 5,
5
(
5
)
xn
n=0
∞
n
n
∑ (−1) (− x)
=
n=0
∞
∑ (−1)
2n
xn =
n=0
∞
∑ xn
n=0
−2
1
1
h( x ) = 2
=
+
=
x −1 1+ x 1− x
∞
n
∑ (−1) x n
+
n=0
∞
∑ xn
=
n=0
∞
∑ ⎡⎣(−1)
n
n=0
+ 1⎤ x n
⎦
∞
= 2 ∑ x 2 n , ( −1, 1) (See Exercise 15.)
= 2 + 0 x + 2 x 2 + 0 x3 + 2 x 4 + 0 x5 + 2 x 6 +
n=0
x
1
1
1 ∞
1 ∞ n
n n
1
x
=
−
=
−
−
(
)
∑
∑x
x2 − 1
2(1 + x)
2(1 − x)
2n=0
2n=0
18. h( x) =
=
1 ∞ ⎡
1
n
∑ (−1) − 1⎤⎦ x n = 2 ⎡⎣0 − 2 x + 0 x2 − 2 x3 + 0 x 4 − 2 x5 +
2n=0 ⎣
=
∞
1 ∞
(−2) x 2 n +1 = − ∑ x 2n +1, (−1, 1)
∑
2n=0
n=0
19. By taking the first derivative, you have
−1
(x
+ 1)
2
=
d ⎡∞
n n⎤
⎢ ∑ ( −1) x ⎥ =
dx ⎣n = 0
⎦
∞
2
+ 1)
3
=
∑ (−1)
n
nx n −1 =
n =1
ln ( x + 1) =
∞
n +1
∑ (−1) (n + 1) x n , (−1, 1).
n=0
d2 ⎡ 1 ⎤
2
. Therefore,
=
dx 2 ⎢⎣ x + 1⎥⎦
( x + 1)3
⎤
d2 ⎡ ∞
d ⎡∞
n
n
n −1 ⎤
−1) x n ⎥ =
⎢∑ (−1) nx ⎥ =
2 ⎢∑ (
dx ⎣n = 0
dx ⎣n =1
⎦
⎦
21. By integrating, you have
⎡
∞
∫ ⎢⎣n∑= 0 (−1)
n
1
∫ x + 1 dx
⎤⎦
d ⎡ 1 ⎤
−1
. Therefore,
=
dx ⎢⎣ x + 1⎥⎦
x
( + 1)2
20. By taking the second derivative, you have
(x
(See Exercise 17.)
∞
n
∑ (−1) n(n − 1) x n − 2
n=2
=
∞
n
∑ (−1) (n + 2)(n + 1) x n , (−1, 1).
n=0
= ln ( x + 1). Therefore,
⎤
x n ⎥ dx = C +
⎦
∞
∑
n=0
(−1)n x n +1 , −1 <
n +1
x ≤ 1.
To solve for C, let x = 0 and conclude that C = 0. Therefore,
ln ( x + 1) =
∞
∑
n=0
(−1)
n
x n +1
n +1
, ( −1, 1].
© 2010 Brooks/Cole, Cengage Learning
Section 9.9
Representation of Functions by Power Series
355
22. By integrating, you have
1
∫1 +
x
1
dx = ln (1 + x) + C1 and
∫1 −
x
dx = −ln (1 − x) + C2 .
f ( x) = ln (1 − x 2 ) = ln (1 + x) − ⎡−
⎣ ln (1 − x)⎤⎦. Therefore,
ln (1 − x 2 ) =
=
1
∫1 +
x
1
∫1 −
dx −
x
⎡∞
n n⎤
∫ ⎣⎢n∑= 0 (−1) x ⎦⎥ dx −
∞
∑
= C +
n=0
dx
⎡
⎡ ∞ n⎤
∫ ⎣⎢n∑= 0 x ⎦⎥ dx = ⎢⎢C1 +
⎣
⎡( −1)n − 1⎤ x n + 1
⎣
⎦
= C +
n +1
∞
∑
n=0
∞
∑
x n +1 ⎤ ⎡
⎥ − ⎢C2 +
n +1 ⎥ ⎣
⎦
(−1)
n=0
n
−2 x 2 n + 2
= C +
2n + 2
∞
∑
∞
x n +1 ⎤
∑ n + 1⎥
n=0
⎦
(−1) x 2n + 2
n +1
n=0
To solve for C, let x = 0 and conclude that C = 0. Therefore,
x 2n + 2
, ( −1, 1).
n +1
∞
ln (1 − x 2 ) = − ∑
n=0
23.
24.
1
=
x +1
2
∞
n
∑ (−1) ( x 2 )
n
=
n=0
∞
∑ (−1)
n
x 2 n , ( −1, 1)
n=0
∞
2x
n
= 2 x ∑ ( −1) x 2 n (See Exercise 23.)
x +1
n=0
2
=
∞
∑ (−1)
n
2 x 2 n +1
n=0
(
)
d
2x
Because
ln ( x 2 + 1) = 2
, you have
dx
x +1
ln ( x 2 + 1) =
⎡∞
n
2 n +1 ⎤
⎢
∫ ⎣n∑= 0 (−1) 2 x ⎥⎦ dx = C +
∞
∑
(−1)
n=0
n
x2n + 2
n +1
, −1 ≤ x ≤ 1.
To solve for C, let x = 0 and conclude that C = 0. Therefore,
ln ( x 2 + 1) =
25. Because,
26. Because
(−1)n x 2 n + 2 ,
∞
∑
n +1
n=0
1
=
x +1
1
∫ 4x2
+1
arctan ( 2 x) = 2 ∫
∞
∑ (−1)
n
[−1, 1].
x n , you have
n=0
dx =
1
4x + 1
2
=
∞
n
∑ (−1) (4 x 2 )
n=0
n
=
∞
∑ (−1)
n
4n x 2 n =
n=0
∞
2n ⎛ 1 1 ⎞
n
∑ (−1) (2 x) , ⎜⎝ − 2 , 2 ⎟⎠.
n=0
1
arctan ( 2 x), you can use the result of Exercise 25 to obtain
2
∞
⎡∞
⎤
(−1) 4n x 2n +1 , − 1 < x < 1 .
n
dx = 2 ∫ ⎢ ∑ ( −1) 4n x 2 n ⎥ dx = C + 2 ∑
2
4x + 1
2n + 1
2
2
n=0
⎣n = 0
⎦
n
1
To solve for C, let x = 0 and conclude that C = 0. Therefore,
∞
arctan ( 2 x) = 2 ∑
n=0
(−1)n 4n x 2n +1 , ⎛ − 1 ,
2n + 1
1⎤
.
⎜
2
2 ⎦⎥
⎝
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
356
Infinite Series
x2
x2
x3
≤ ln ( x + 1) ≤ x −
+
2
2
3
27. x −
5
S3
f
−4
8
S2
−3
x
0.0
0.2
0.4
0.6
0.8
1.0
0.000
0.180
0.320
0.420
0.480
0.500
0.000
0.182
0.336
0.470
0.588
0.693
0.000
0.183
0.341
0.492
0.651
0.833
2
x
2
S2 = x −
ln ( x + 1)
2
3
x
x
+
2
3
S3 = x −
x2
x3
x4
x2
x3
x4
x5
+
−
≤ ln ( x + 1) ≤ x −
+
−
+
2
3
4
2
3
4
5
28. x −
5
S5
f
−4
8
S4
−3
x
2
3
x
x
x
+
−
2
3
4
S4 = x −
ln ( x + 1)
2
29.
∞
∑
3
(−1)n +1 ( x
− 1)
n
n
n =1
(a)
4
0.2
0.4
0.6
0.8
1.0
0.0
0.18227
0.33493
0.45960
0.54827
0.58333
0.0
0.18232
0.33647
0.47000
0.58779
0.69315
0.0
0.18233
0.33698
0.47515
0.61380
0.78333
5
x
x
x
x
+
−
+
2
3
4
5
S5 = x −
0.0
4
=
(x
− 1)
1
−
(x
− 1)
2
2
+
(x
− 1)
3
3
−
3
n=3
n=1
0
4
n=2
n=6
−3
(b) From Example 4,
∞
∑
(−1)n +1 ( x
− 1)
n
n
n =1
=
∞
∑
(−1)n ( x
− 1)
n +1
n=0
n +1
= ln x, 0 < x ≤ 2, R = 1.
(c) x = 0.5:
∞
∑
n =1
(−1) (−1 2)
n +1
n
n
=
∞
∑
−(1 2)
n =1
n
n
≈ −0.693147
⎛1⎞
(d) This is an approximation of ln ⎜ ⎟. The error is approximately 0. [The error is less than the first omitted term,
⎝ 2⎠
1 (51 ⋅ 251 ) ≈ 8.7 × 10−18 . ]
© 2010 Brooks/Cole, Cengage Learning
Section 9.9
30.
(−1)
∞
n
x 2n +1
∑
n = 0 ( 2 n + 1)!
Matches (c)
4
32. g ( x) = x −
−6
6
( −1)
n
x
2 n +1
(b)
∑
n = 0 ( 2n + 1)!
(c)
∑ (2n + 1)!
n=0
33. g ( x) = x −
= sin x, R = ∞
( −1)n (1 2)2 n +1
∞
34. g ( x) = x −
∑ ( −1)
n
n=0
35. arctan
1
=
4
Because
∞
n
∑ (−1)
n=0
2n + 1
∞
∑ (−1)
n
n=0
34
(−1)n
∑
2 n +1
n = 0 ( 2 n + 1)4
∞
=
1
1
1
−
+
+
4 192
5120
∞
x4n + 2
2n + 1
x4n + 3
dx =
∑ (−1) (4n + 3)(2n + 1)
n=0
arctan x 2 dx =
∑ (−1) (4n + 3)(2n + 1)
n=0
∫ arctan x
∫0
=
x 2 n +1
.
2n + 1
1
1
1
1
< 0.001, you can approximate the series by its first two terms: arctan ≈
−
≈ 0.245.
5120
4
4 192
arctan x 2 =
36.
(1 4)
2 n +1
x3
x5
x7
,
+
−
3
5
7
Matches (b)
⎛1⎞
(d) This is an approximation of sin ⎜ ⎟. The error is
⎝ 2⎠
approximately 0.
∞
x3
x5
+
3
5
Matches (a)
≈ 0.4794255386
In Exercises 35–38, arctan x =
x3
, cubic with 3 zeros.
3
Matches (d)
−4
∞
357
31. g ( x) = x line
x3
x5
+
−
3!
5!
= x −
(a)
Representation of Functions by Power Series
2
=
∞
n
n
( 3 4) 4 n + 3
+ C, C = 0
=
∞
34 n + 3
∑ (−1) (4n + 3)(2n + 1)44n + 3
n=0
n
27
2187
177,147
−
+
192 344,064
230,686,720
Because 177,147 230,686,720 < 0.001, you can approximate the series by its first two terms: 0.134.
arctan x 2
1 ∞
n (x )
= ∑ ( −1)
=
x
x n=0
2n + 1
2
37.
∫
12
∫0
2
∞
2 n +1
2
n=0
arctan x
dx =
x
n=0
arctan x 2
dx =
x
∑ (−1) (4n + 2)(2n + 1)24n + 2
n=0
Because
x
4n + 2
∞
∑ (−1)
∑ (−1) (4n + 2)(2n + 1)
∞
n
n
1
x 4n +1
2n + 1
+ C ( Note: C = 0)
=
1
1
−
+
8 1152
1
< 0.001, you can approximate the series by its first term:
1152
12
∫0
arctan x 2
dx ≈ 0.125.
x
© 2010 Brooks/Cole, Cengage Learning
358
Chapter 9
Infinite Series
x 2 arctan x =
38.
∞
∑ (−1)
n
n=0
∫x
12
∫0
2
arctan x dx =
x 2 arctan x dx =
In Exercises 39–42, use
40.
41.
1
(1 − x)
x2n + 4
∑ (−1) (2n + 4)(2n + 1)
n=0
n
∞
1
∑ (−1) (2n + 4)(2n + 1)22n + 4
n=0
n
=
1
1
−
+
64 1152
1
< 0.001, you can approximate the series by its first term:
1152
Because
39.
∞
x2n + 3
2n + 1
∞
x
= x ∑ nx n −1 =
(1 − x)2
n =1
1+ x
=
(1 − x)2
∞
∑ xn , x
=
1
(1 − x)2
+
< 1.
44. (a)
n=0
d ⎡ 1 ⎤
d ⎡ ∞ n⎤
=
⎢∑ x ⎥ =
⎢
⎥
dx ⎣1 − x ⎦
dx ⎣n = 0 ⎦
=
2
1
=
1− x
∞
∑ nx n , x
∞
∑ nx n −1, x < 1
n =1
42.
(1 − x)
2
(1 − x)2
∑ n( x n −1 + x n ),
∞
∑ (2n + 1) x n ,
x <1
∞
n =1
=
2 ∞ ⎛ 2⎞
∑ n⎜ ⎟
9 n =1 ⎝ 3 ⎠
n −1
=
1 ∞ ⎛9⎞
9 ∞ ⎛9⎞
n⎜ ⎟ =
∑
∑ n⎜ ⎟
10 n =1 ⎝ 10 ⎠
100 n =1 ⎝ 10 ⎠
46. Because
x <1
n
n=0
2
1
= 2
9 ⎡1 − ( 2 3)⎤ 2
⎣
⎦
n −1
9
1
⋅
= 9
100 ⎡1 − (9 10)⎤ 2
⎣
⎦
1
1
, substitute ( − x) into the
=
1+ x
1 − ( − x)
∞
1
1
, substitute ( x 2 ) into the
=
1 − x2
1 − ( x2 )
geometric series.
∑ (2n + 1) x
n +1
, x <1
n=0
⎛
⎞
1
1
= 5⎜
⎟⎟, substitute ( − x) into the
⎜
1+ x
⎝ 1 − (− x) ⎠
geometric series and then multiply the series by 5.
47. Because
1
dx, integrate the series
1− x
and then multiply by ( −1).
48. Because ln (1 − x ) = − ∫
n
∑ nP(n)
=
geometric series.
(See Exercise 41.)
E ( n) =
n
n
(b)
45. Because
= x ∑ ( 2n + 1) x =
⎛1⎞
43. P( n) = ⎜ ⎟
⎝ 2⎠
( 23 )
<1
x
∞
∞
1∞
∑n
3 n =1
n =1
n=0
x(1 + x)
x 2 arctan x dx ≈ 0.016.
=
n =1
=
12
∫0
=
∞
⎛1⎞
∑ n⎜⎝ 2 ⎟⎠
n
=
n =1
1 ∞ ⎛1⎞
∑ n⎜ ⎟
2 n =1 ⎝ 2 ⎠
n −1
1
1
= 2
2 ⎡1 − (1 2)⎤ 2
⎣
⎦
Because the probability of obtaining a head on a single
1
toss is , it is expected that, on average, a head will be
2
obtained in two tosses.
49. Let arctan x + arctan y = θ . Then,
tan (arctan x + arctan y ) = tan θ
tan (arctan x) + tan (arctan y )
= tan θ
1 − tan (arctan x) tan (arctan y )
x+ y
= tan θ
1 − xy
⎛x + y⎞
arctan ⎜
⎟ = θ.
⎝ 1 − xy ⎠
Therefore,
⎛x + y⎞
arctan x + arctan y = arctan ⎜
⎟ for xy ≠ 1.
⎝ 1 − xy ⎠
50. (a) From Exercise 49, you have
arctan
⎡ (120 119) + ( −1 239) ⎤
120
1
120
⎛ 1 ⎞
− arctan
= arctan
+ arctan ⎜ −
⎥
⎟ = arctan ⎢
119
239
119
⎝ 239 ⎠
⎣⎢1 − (120 119)( −1 239) ⎦⎥
π
⎛ 28,561 ⎞
= arctan ⎜
⎟ = arctan 1 =
28,561
4
⎝
⎠
© 2010 Brooks/Cole, Cengage Learning
Section 9.9
(b) 2 arctan
Representation of Functions by Power Series
359
⎡ 2(1 5) ⎤
1
1
1
10
5
⎥ = arctan
= arctan + arctan = arctan ⎢
= arctan
2
5
5
5
24
12
⎢⎣1 − (1 5) ⎥⎦
4 arctan
⎡ 2(5 12) ⎤
1
1
1
5
5
120
⎥ = arctan
= 2 arctan + 2 arctan = arctan
+ arctan
= arctan ⎢
2
5
5
5
12
12
119
⎣⎢1 − (5 12) ⎦⎥
4 arctan
1
1
120
1
π
=
− arctan
= arctan
− arctan
(see part (a ).)
5
239
119
239
4
1 ⎤
⎡ 1
⎢ 2 + 2 ⎥
1
1
1
4
= arctan
51. (a) 2 arctan = arctan + arctan = arctan ⎢
2⎥
2
2
2
3
1
1
2
−
( )⎥
⎢
⎣
⎦
2 arctan
⎡ ( 4 3) − (1 7) ⎤
1
1
4
25
π
⎛ 1⎞
− arctan = arctan + arctan ⎜ − ⎟ = arctan ⎢
= arctan 1 =
⎥ = arctan
2
7
3
25
4
⎝ 7⎠
⎢⎣1 + ( 4 3)(1 7) ⎥⎦
(b) π = 8 arctan
52. (a) arctan
5
7
3
5
7
⎡ 1 (0.5)3
1
1
(0.5) − (0.5) ⎤⎥ − 4⎡⎢ 1 − (1 7) + (1 7) − (1 7) ⎤⎥ ≈ 3.14
− 4 arctan ≈ 8⎢ −
+
2
7
3
5
7 ⎥
3
5
7 ⎥
⎢⎣ 2
⎢⎣ 7
⎦
⎦
⎡ (1 2) + (1 3) ⎤
⎛5 6⎞
π
1
1
+ arctan = arctan ⎢
⎥ = arctan ⎜
⎟ =
−
2
3
1
1
2
1
3
5
6
4
(
)(
)
⎝
⎠
⎣⎢
⎦⎥
1
1⎤
⎡
(b) π = 4 ⎢arctan + arctan ⎥
2
3⎦
⎣
⎡1
(1 2)3 + (1 2)5 − (1 2)7 ⎤⎥ + 4⎡⎢1 − (1 3)3 + (1 3)5 − (1 3)7 ⎤⎥ ≈ 4 0.4635 + 4 0.3217 ≈ 3.14
= 4⎢ −
(
) (
)
3
5
7 ⎥
3
5
7 ⎥
⎢⎣ 2
⎢⎣ 3
⎦
⎦
53. From Exercise 21, you have
ln ( x + 1) =
∞
∑
(−1)n x n +1
n +1
n=0
=
(−1)
∞
∑
So,
∞
∑ (−1)
n =1
n +1
∞
(−1)n −1 x n
n =1
n
∑
n +1 n
x
n
n =1
=
1
=
2n n
∑ (−1)
n
1
=
2n + 1
∑
(−1) (1 2)
n +1
n
n =1
n +1
n =1
1
=
3n n
∞
∑
∞
∑ (−1)
n=0
(−1) (1 3)
n +1
∑ (−1)
n =1
2n
=
5n n
∑
n =1
π
n
2
2n +1
1
=
(2n + 1)
∞
∑ (−1)
(−1) (2 5)
n
(1 2)2 n +1
2n + 1
n=0
∞
∑ (−1)
n =1
n +1
1
=
2 n −1
3
(2n − 1)
=
n +1
≈ 0.7854
4
1
≈ 0.4636.
2
58. From Exercise 56, you have
n
n =1
∞
2n + 1
n=0
n
55. From Exercise 53, you have
n +1
(1)
2n +1
= arctan
4
⎛1
⎞
= ln ⎜ + 1⎟ = ln ≈ 0.2877.
3
⎝3
⎠
∞
n
x 2 n +1
.
2n + 1
57. From Exercise 56, you have
54. From Exercise 53, you have
∞
∑ (−1)
n
3
⎛1
⎞
= ln ⎜ + 1⎟ = ln ≈ 0.4055.
2
2
⎝
⎠
∑ (−1)
∞
= arctan 1 =
∞
n
n=0
∞
n=0
.
∞
∑ (−1)
56. From Example 5, you have arctan x =
∞
∑ (−1)
= arctan
n
n
2 n +1
3
n=0
n=0
n
7
⎛2
⎞
= ln ⎜ + 1⎟ = ln ≈ 0.3365.
5
⎝5
⎠
∞
∑ (−1)
n
1
(2n + 1)
(1 3)2n +1
2n + 1
1
≈ 0.3218.
3
59. f ( x) = arctan x is an odd function (symmetric to the
origin).
© 2010 Brooks/Cole, Cengage Learning
360
Chapter 9
Infinite Series
60. The approximations of degree
3, 7, 11, …, ( 4n − 1, n = 1, 2, …) have relative extrema.
65.
∞
∑ nan x
n −1
n=0
, the radius of
n=0
n=0
∞
⎛ x⎞
n
n=0
x⎞
⎛
= (1 + 1) + ⎜ x + ⎟ +
5
⎝
⎠
66.
1⎞
1⎞ ⎛
1⎞
⎛
⎛
∑ ⎜⎝1 + 5 ⎟⎠ x n = ⎜⎝1 + 5 ⎟⎠ + ⎜⎝1 + 5 ⎟⎠ x +
n=0
∞
n
∞
⎛ x⎞
∑ x + ∑ ⎜⎝ 5 ⎟⎠ =
n=0
n=0
∑
n=0
∞
(1
3
3∑
(−1)
n
∞
)
2 n +1
⎛ 1 ⎞
3 arctan ⎜
⎟
⎝ 3⎠
=
⎛π ⎞
3 ⎜ ⎟ ≈ 0.9068997
⎝6⎠
(−1)n π 2 n +1
=
∞
x 2n +1
.
2n + 1
2n + 1
(π 3)2n +1
∑ (−1) (2n + 1)!
n=0
n
⎛π ⎞
= sin ⎜ ⎟ =
⎝3⎠
The formula should be
n
+ 1)
n
3
3
=
∑ 32n +1(2n + 1)!
n=0
∞
∞
(
n=0
64. You can verify that the statement is incorrect by
calculating the constant terms of each side:
∑ ⎜⎝ 5 ⎟⎠
n=0
(−1)
2n
3 ) ( 2n
n
∞
∑
=
perhaps at the endpoints, x = 3 and x = −5. )
+
=
n =1
63. Because the first series is the derivative of the second
series, the second series converges for x + 1 < 4 (and
∞
(−1)
n
∑ 3n (2n + 1)
convergence is the same, 3.
∑ xn
∞
∑ (−1)
From Example 5 you have arctan x =
∞
∞
n
n=0
61. The series in Exercise 56 converges to its sum at a
slower rate because its terms approach 0 at a much
slower rate.
d ⎡∞
n⎤
62. Because
⎢ ∑ an x ⎥ =
dx ⎣n = 0
⎦
( −1)
∑ 3n (2n + 1)
3
≈ 0.866025
2
67. Using a graphing utility, you obtain the following partial
sums for the left hand side. Note that
1 π ≈ 0.3183098862.
n
⎡
⎛1⎞ ⎤
⎢1 + ⎜ ⎟ ⎥ x n .
⎝ 5 ⎠ ⎦⎥
⎣⎢
n = 0: S0 ≈ 0.3183098784
n = 1: S1 ≈ 0.3183098862
Section 9.10 Taylor and Maclaurin Series
1. For c = 0, you have:
f ( x) = e2 x
f ( n ) ( x ) = 2 n e 2 x ⇒ f ( n ) ( 0) = 2 n
4 x2
8 x3 16 x 4
+
+
+
2!
3!
4!
e2 x = 1 + 2 x +
=
∞
∑
n=0
( 2 x)n .
n!
2. For c = 0, you have:
f ( x ) = e3 x
f (n) ( x) = 3n e3 x ⇒ f (n) (0) = 3n
e3 x = 1 + 3 x +
9 x2
27 x3
+
+
2!
3!
=
∞
∑
n=0
(3 x ) n .
n!
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
Taylor and Maclaurin Series
361
3. For c = π 4, you have:
f ( x) = cos( x)
⎛π ⎞
f⎜ ⎟ =
⎝4⎠
f ′( x) = −sin ( x)
2
⎛π ⎞
f ′⎜ ⎟ = −
4
2
⎝ ⎠
f ′′( x) = −cos( x)
2
⎛π ⎞
f ′′⎜ ⎟ = −
2
⎝4⎠
f ′′′( x) = sin ( x)
⎛π ⎞
f ′′′⎜ ⎟ =
⎝4⎠
2
2
f (4) ( x) = cos( x)
⎛π ⎞
f (4) ⎜ ⎟ =
⎝4⎠
2
2
2
2
and so on. Therefore, you have:
cos x =
∞
∑
f (n) (π 4) ⎣⎡ x − (π 4)⎤⎦
n
n!
n=0
=
⎡ x − (π 4)⎤⎦
⎡ x − (π 4)⎤⎦
2 ⎡⎢
π ⎞ ⎡ x − (π 4)⎤⎦
⎛
1 − ⎜x − ⎟ − ⎣
+ ⎣
+ ⎣
−
2 ⎢
4
2!
3!
4!
⎝
⎠
⎣
=
2 ∞ ( −1)
∑
2 n=0
2
[Note: ( −1)
n( n + 1) 2
n( n + 1) 2
3
4
⎤
⎥
⎥
⎦
n
⎡⎣ x − (π 4)⎤⎦
.
n!
= 1, −1, −1, 1, 1, −1, −1, 1, … ]
4. For c = π 4, you have:
f ( x) = sin x
⎛π ⎞
f⎜ ⎟ =
⎝4⎠
2
2
f ′( x) = cos x
⎛π ⎞
f ′⎜ ⎟ =
⎝4⎠
2
2
f ′′( x) = −sin x
2
⎛π ⎞
f ′′⎜ ⎟ = −
2
⎝4⎠
f ′′′( x) = −cos x
2
⎛π ⎞
f ′′′⎜ ⎟ = −
4
2
⎝ ⎠
f (4) ( x) = sin x
⎛π ⎞
f (4) ⎜ ⎟ =
⎝4⎠
2
2
and so on. Therefore you have:
sin x =
∞
∑
n=0
f (n) (π 4) ⎡⎣ x − (π 4)⎤⎦
n!
n
=
⎡ x − (π 4)⎤⎦
⎡ x − (π 4)⎤⎦
2 ⎡⎢
π ⎞ ⎡ x − (π 4)⎤⎦
⎛
1 + ⎜x − ⎟ − ⎣
− ⎣
+ ⎣
+
2 ⎢
4⎠
2!
3!
4!
⎝
⎣
=
2 ⎧⎪ ∞ ( −1)
⎨∑
2 ⎪n = 0
⎩
2
n( n − 1) 2
⎡⎣ x − (π 4)⎤⎦
n
( + 1)!
n +1
3
4
⎤
⎥
⎥
⎦
⎫
⎪
+ 1⎬.
⎪⎭
© 2010 Brooks/Cole, Cengage Learning
362
Chapter 9
Infinite Series
6. For c = 2, you have
5. For c = 1, you have
1
= x −1
x
f ( x) =
1
−1
= (1 − x)
1− x
f (1) = 1
f ( x) =
f ′( x) = − x −2
f ′(1) = −1
f ′( x) = (1 − x)
f ′′( x) = 2 x
f ′′(1) = 2
f ′′( x) = 2(1 − x)
−3
f ′′′( x) = 6(1 − x)
−4
−3
f ′′′( x) = −6 x
f ′′′(1) = −6
−4
and so on. Therefore, you have
1
=
x
∞
f (n) (1)( x − 1)
n=0
n!
∑
f ′′( 2) = −2
f ′′′( 2) = 6
∞ f ( n) 2 x − 2
1
( )(
)
= ∑
1− x
n
!
n=0
2
3
= −1 + ( x − 2) − ( x − 2) + ( x − 2) −
n
6( x − 1)
2
−
2!
3
+
3!
=
= 1 − ( x − 1) + ( x − 1) − ( x − 1) +
2
=
f ′( 2) = 1
−2
and so on. Therefore you have
n
2( x − 1)
= 1 − ( x − 1) +
f ( 2) = −1
3
∞
n +1
n
∑ (−1) ( x − 2)
n=0
∞
n
n
∑ (−1) ( x − 1)
n=0
7. For c = 1, you have,
f ( x) = ln x
f (1) = 0
1
x
f ′( x) =
f ′(1) = 1
1
x2
2
f ′′′( x) = 3
x
6
4)
(
f ( x) = − 4
x
24
f (5) ( x) = 5
x
f ′′( x) = −
f ′′(1) = −1
f ′′′(1) = 2
f (4) (1) = −6
f (5) (1) = 24
and so on. Therefore, you have:
ln x =
∞
f (n) (1)( x − 1)
n=0
n!
∑
= 0 + ( x − 1) −
= ( x − 1) −
=
∞
n
∑ (−1)
(x
(x
(x
− 1)
2!
− 1)
2
+
2
− 1)
2
+
(x
2( x − 1)
3
3!
− 1)
3
3
−
(x
−
6( x − 1)
− 1)
4
4!
4
+
(x
4
+
− 1)
5
24( x − 1)
5!
5
−
5
−
n +1
n +1
n=0
n
.
8. For c = 1, you have:
f ( x) = e x
f (n) ( x) = e x ⇒ f (n) (1) = e
ex =
∞
f (n) (1)( x − 1)
n=0
n!
∑
n
⎡
( x − 1)2 + ( x − 1)3 + ( x − 1)4 +
= e ⎢1 + ( x − 1) +
2!
3!
4!
⎣⎢
∞
⎤
( x − 1)n .
⎥ = e∑
n!
n=0
⎦⎥
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
Taylor and Maclaurin Series
363
9. For c = 0, you have
f ( x) = sin 3x
f (0) = 0
f ′( x) = 3 cos 3x
f ′(0) = 3
f ′′( x) = −9 sin 3 x
f ′′(0) = 0
f ′′′( x) = −27 cos 3 x
f ′′′(0) = −27
f (4) ( x) = 81 sin 3 x
f (4) (0) = 0
and so on. Therefore you have
sin 3 x =
∞
∑
n=0
f ( n ) ( 0) x n
27 x3
= 0 + 3x + 0 −
+0+
n!
3!
=
(−1) (3x)
∑ (2n + 1)!
n=0
∞
n
2 n +1
10. For c = 0, you have.
f ( x) = ln ( x 2 + 1)
f ′( x) =
2x
x2 + 1
f ′′( x) =
2 − 2x2
f ′′′( x)
f ( 4) ( x )
f (5) ( x)
f ( 6) ( x )
f ( 0) = 0
f ′(0) = 0
( x 2 + 1)
4 x( x 2 − 3)
=
3
( x 2 + 1)
12( − x 4 + 6 x 2 − 1)
=
4
( x 2 + 1)
48 x( x 4 − 10 x 2 + 5)
=
5
( x 2 + 1)
−240(5 x 6 − 15 x 4 + 15 x 2
=
6
( x 2 + 1)
f ′′(0) = 2
2
f ′′′(0) = 0
f (4) (0) = −12
f (5) (0) = 0
− 1)
f (6) (0) = 240
and so on. Therefore, you have:
ln ( x 2 + 1) =
∞
f (n) (0) x n
n=0
n!
∑
= x2 −
= 0 + 0x +
x4
x6
+
−
2
3
=
∞
∑
2x2
0 x3 12 x 4
0 x5
240 x 6
+
−
+
+
+
2!
3!
4!
5!
6!
(−1)
n=0
n
x2n + 2
n +1
.
11. For c = 0, you have:
f ( x) = sec( x)
f (0) = 1
f ′( x) = sec( x) tan ( x)
f ′(0) = 0
f ′′( x) = sec ( x) + sec( x) tan ( x)
3
f ′′(0) = 1
2
f ′′′( x) = 5 sec ( x) tan ( x) + sec( x) tan ( x)
3
f
( 4)
( x)
f ′′′(0) = 0
3
= 5 sec5 ( x) + 18 sec3 ( x) tan 2 ( x) + sec( x) tan 4 ( x)
sec( x) =
∞
f (n) (0) x n
n=0
n!
∑
=1+
x2
5x4
+
+
2!
4!
f (4) (0) = 5
.
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
364
Infinite Series
12. For c = 0, you have;
f ( x) = tan ( x)
f (0) = 0
f ′( x) = sec 2 ( x)
f ′(0) = 1
f ′′( x) = 2 sec 2 ( x) tan ( x)
f ′′(0) = 0
f ′′′( x) = 2 ⎡⎣sec ( x) + 2 sec ( x) tan ( x)⎤⎦
f ′′′(0) = 2
f (4) ( x) = 8⎡⎣sec 4 ( x) tan ( x) + sec 2 ( x) tan 3 ( x)⎤⎦
f (4) (0) = 0
f (5) ( x) = 8⎡⎣2 sec6 ( x) + 11 sec 4 ( x) tan 2 ( x) + 2 sec 2 ( x) tan 4 ( x)⎤⎦
f (5) (0) = 16
4
tan ( x) =
2
∞
f ( n ) ( 0) x n
n=0
n!
∑
2
= x +
2 x3 16 x5
+
+
3!
5!
= x +
x3
2 5
+
x +
3
15
.
(−1)n x 2 n .
∑ (2n)!
n=0
∞
13. The Maclaurin series for f ( x) = cos x is
Because f (n + 1) ( x) = ±sin x or ± cos x, you have f (n + 1) ( z ) ≤ 1 for all z. So by Taylor's Theorem,
f (n + 1) ( z )
0 ≤ Rn ( x) =
Because lim
n→∞
(n
x
(n
+ 1)!
n +1
+ 1)!
x n +1 ≤
x
(n
n +1
.
+ 1)!
= 0, it follows that Rn ( x) → 0 as n → ∞. So, the Maclaurin series for cos x converges to
cos x for all x.
14. The Maclaurin series for f ( x) = e −2x is
(−2 x)n .
∞
∑
n!
n=0
f
(n + 1)
( x)
= ( −2)
0 ≤ Rn( x) =
Because lim
n→∞
n + 1 −2 x
e
. So, by Taylor's Theorem,
f (n + 1) ( z )
(n
+ 1)!
x n +1 =
(−2)n +1 x n +1
(n + 1)!
(−2)n +1 e−2 z x n +1 .
(n + 1)!
= lim
n→∞
( 2 x)n + 1
(n + 1)!
= 0, it follows that Rn ( x) → 0 as n → ∞.
So, the Maclaurin Series for e −2 x converges to e −2 x for all x.
15. The Maclaurin series for f ( x) = sinh x is
∞
x 2n +1
∑ (2n + 1)!.
n=0
f (n + 1) ( x) = sinh x (or cosh x). For fixed x,
0 ≤ Rn ( x) =
f (n + 1) ( z )
(n
+ 1)!
x n +1 =
sinh ( z )
(n
+ 1)!
x n + 1 → 0 as n → ∞.
(The argument is the same if f (n + 1) ( x) = cosh x ). So, the Maclaurin series for sinh x converges to sinh x for all x.
16. The Maclaurin series for f ( x) = cosh x is
f
(n + 1)
( x)
∞
x2n
∑ (2n)!.
n=0
= sinh x (or cosh x). For fixed x,
0 ≤ Rn ( x) =
f (n + 1) ( z )
(n
+ 1)!
x n +1 =
sinh ( z )
(n
+ 1)!
x n + 1 → 0 as n → ∞.
(The argument is the same if f (n + 1) ( x) = cosh x ). So, the Maclaurin series for cosh x converges to cosh x for all x.
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
17. Because (1 + x)
(1 + x)
−2
−k
k ( k + 1) x 2
= 1 − kx +
2(3) x 2
= 1 − 2x +
2!
2!
2(3)( 4) x3
−
3!
−
+
k ( k + 1)( k + 2) x3
+
3!
2(3)( 4)(5) x 4
4!
−
Taylor and Maclaurin Series
365
, you have
= 1 − 2 x + 3x 2 − 4 x3 + 5 x 4 −
∞
n
∑ (−1) (n + 1) x n .
=
n=0
18. Because (1 + x)
−k
k ( k + 1) x 2
= 1 − kx +
2!
−
k ( k + 1)( k + 2) x3
3!
+
you have
(1 + x)
−4
4(5) 2 4(5)(6) 3 4(5)(6)(7) 4
x −
x +
x = 1 − 4 x + 10 x 2 − 20 x3 + 35 x 4 −
2!
3!
4!
= 1 − 4x +
19. Because (1 + x)
⎣⎡1 + ( − x)⎤⎦
−k
k ( k + 1) x 2
= 1 − kx +
2!
(1)(3) x 2 + (1)(3)(5) x3 +
x
+
2
22 2!
23 3!
=1+
∑
20. Because (1 + x)
∞
−1 2
−k
1
2 n!
4 + x2
22.
k ( k + 1) x 2
2!
+
you have
+ 3)! n
x
3!n!
k ( k + 1)( k + 2) x3
3!
(1 2)(3 2) x 4 − (1 2)(3 2)(5 2) x6 +
1 2
x +
2
2!
3!
1)(3) 4
1)(3)(5) 6
(
(
1 2
x +
=1− x + 2 x −
2
2 2!
23 3!
∞
∑
1⎡
⎢1 +
2 ⎢⎣
∞
∑
( 2n
1⋅3⋅5
− 1)
n
2 n!
x 2n
−1 2
and because (1 + x)
−1 2
=1+
∞
∑
(−1)n1 ⋅ 3 ⋅ 5
( 2n
n
2 n!
n =1
2n
− 1)( x 2) ⎤
1
⎥ =
+
2
⎥⎦
− 1) x n
2 n!
n =1
(−1)n1 ⋅ 3 ⋅ 5
( 2n
n
∞
∑
(−1)n1 ⋅ 3 ⋅ 5
2
n =1
( 2n
3n + 1
, you have
− 1) x 2 n
n!
.
−3
3
3
=
3( 4)(5) ⎛ x ⎞
1 ⎧⎪
⎛ x ⎞ 3( 4) ⎛ x ⎞
⎨1 − 3⎜ ⎟ +
⎜ ⎟ −
⎜ ⎟ +
8 ⎩⎪
2
2!
2
3! ⎝ 2 ⎠
⎝ ⎠
⎝ ⎠
k = −3
2
1
23.
−
1⎛
x⎞
⎜1 + ⎟ ,
8⎝
2⎠
+ x)
(2 + x)
, you have
n=0
(n
.
=
1
(2
+
n
=1−
2
⎛ 1 ⎞⎡
⎛ x⎞ ⎤
= ⎜ ⎟ ⎢1 + ⎜ ⎟ ⎥
⎝ 2 ⎠ ⎣⎢
⎝ 2 ⎠ ⎥⎦
=
− 1) x n
n
n =1
4 + x2
( 2n
1⋅3⋅5
= 1 − kx +
=1+
1
3!
=1+
n =1
21.
k ( k + 1)( k + 2) x3
∞
∑ (−1)
(1 2)(3 2) x 2 + (1 2)(3 2)(5 2) x3 +
⎛1⎞
= 1 + ⎜ ⎟x +
2!
3!
⎝ 2⎠
−1 2
⎡1 + ( − x 2 )⎤
⎣
⎦
−
=
1 + x = (1 + x) ,
12
1+ x =1+
3
⎫⎪
1⎡
⎬ = ⎢1 +
8
⎣
⎭⎪
∞
(n
∑ (−1)
n
1
x +
2
∞
n =1
+ 2)! n ⎤
x ⎥
2n +1 n! ⎦
k =12
1 2( −1 2) 2 1 2( −1 2)( −3 2) 3
1
x +
x +
x +
2
2!
3!
=1+
∑ (−1)
n=2
n +1
( 2n
1⋅3⋅5
n
2 n!
− 3)
xn
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
366
Infinite Series
(1 4)(−3 4) x 2 + (1 4)(−3 4)(−7 4) x3 +
1
x +
4
2!
3!
1
3 2
3 ⋅ 7 3 3 ⋅ 7 ⋅ 11 4
=1+ x − 2 x + 3 x −
x +
4
4 2!
4 3!
44 4!
24. (1 + x)
14
=1+
25. Because (1 + x)
12
12
26. Because (1 + x)
12
12
∞
n=0
e
x2 2
( x 2 2)
∞
∑
=
ex =
∞
n
xn
∑ n!
∞
29. ln x =
n=0
n!
∞
∑ (−1)
=
n −1
(x
∞
∑
ln ( x + 1) =
(−1)
(−1) ( x
n −1
ln ( x + 1) =
2
32.
− 1)
− 3) x n
∞
∑
(−1)
n +1
1⋅3⋅5
( 2n
− 3) x3n
n
2 n!
n=2
.
=1+
∞
∑
,
x
− 1)
,
(−1)
∞
(−1)n (3x)2n +1
∞
(−1)n x 2 n +1
∞
(−1)n (π x)2 n +1
2 n +1
sin 3 x =
∑ (2n + 1)!
n=0
0 < x ≤ 2
33.
cos x =
cos 4 x =
∞
−1 < x ≤ 1
cos x =
cos π x =
=1−
(−1)n (4 x)2n
(2n)!
n=0
∞
∑
0 < x ≤ 2
34.
(−1)n x 2n
∑ (2n)!
n=0
=1−
∞
x
9x2
27 x3
81x 4
243 x5
−
+
−
+
2!
3!
4!
5!
n
n
n
= 1 − 3x +
, −1 < x ≤ 1
(−1)n −1 x 2 n ,
n =1
x2
x4
x6
x8
+ 2 + 3 + 4 +
2
2 2! 2 3! 2 4!
n
n −1 n
∑
n = 0 ( 2n + 1)!
sin π x =
( 2n
.
2n n!
n!
sin x =
sin x =
− 3) x 2 n
2n n!
n=2
n=0
n
n =1
31.
1⋅3⋅5
(−1)n +11 ⋅ 3 ⋅ 5
(−1)n 3n x n
n
n =1
∑
( 2n
n +1
n=2
∞
∑
n
n =1
30. ln x =
(−1)
x2
x3
x4
x5
+
+
+
+
2!
3!
4!
5!
= 1+ x +
(−3x)n
∞
∞
∑
− 3) x n
2 n!
n=0
∞
∑
1⋅3⋅5
xn
n
∑
∞
x3
+
2
( 2n
n +1
n=2
x2n
n=0
e −3 x =
(−1)
∞
∑
∑ 2n n!
=
n!
n=0
28.
− 5)
xn
x2
x3
x4
x5
=1+ x +
+
+
+
+
2!
3!
4!
5!
n!
∑
ex =
=1+
( 4n
3 ⋅ 7 ⋅ 11
4n n!
x2
+
2
x
+
2
=1+
n +1
n=2
=1+
you have (1 + x3 )
27.
∑
x
+
2
=1+
you have (1 + x 2 )
(−1)
∞
1
x +
4
=1+
∞
=
x2
x4
x6
+
−
+
2!
4!
6!
(−1)n 42 n x 2 n
(2n)!
n=0
∞
∑
16 x 2
256 x 4
+
−
2!
4!
(−1)n x 2n
∑ (2n)!
n=0
(−1)n (π x)2 n
(2n)!
n=0
∞
∑
∑
n = 0 ( 2 n + 1)!
∑ (2n + 1)!
n=0
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
35.
cos x =
cos x3 2 =
∞
(−1)n x 2 n
∞
(−1) ( x3 2 )
∑ (2n)!
n=0
n
∑
(−1)
∞
sin x =
2 sin x =
3
37.
2n
x
n
x
∑
n = 0 ( 2 n + 1)!
n
⎞
⎟
⎠
e− x = 1 − x +
x2
x3
−
+
2!
3!
e x + e− x = 2 +
=
∞
x 2n +1
∑ (2n + 1)!
n=0
2x2
2 x4
+
+
2!
4!
2 cos h( x) = e x + e− x =
∞
x2n
∑ 2 (2n)!
n=0
1
⎡1 + cos( 2 x)⎤⎦
2⎣
( 2 x ) + ( 2 x ) − ( 2 x) −
1⎡
⎢1 + 1 −
2⎢
2!
4!
6!
⎣
4
6
40. The formula for the binomial series gives (1 + x)
(−1)n (2 x)2 n ⎤⎥
(2n)! ⎦⎥
n=0
⎤
1⎡
⎥ = ⎢1 +
2⎢
⎦⎥
⎣
−1 2
=1+
∞
∑
(−1)n1 ⋅ 3 ⋅ 5
∞
∑
(1 + x 2 )
−1 2
=1+
∞
∑
(−1)n1 ⋅ 3 ⋅ 5
2 n!
)
∫
− 1) x n
, which implies that
− 1) x 2 n
n
2 n!
n =1
x2 + 1 =
( 2n
( 2n
n
n =1
1
x +1
2
dx
= x +
(−1) 1 ⋅ 3 ⋅ 5 ( 2n − 1) x 2n +1
∑
2n ( 2n + 1)n!
n =1
= x −
x3
1 ⋅ 3 x5
1 ⋅ 3 ⋅ 5x7
+
−
+
2⋅3 2⋅4⋅5 2⋅4⋅6⋅7
n
∞
⎛
x3
x5
41. x sin x = x⎜ x −
+
−
3!
5!
⎝
⎞
x4
x6
2
+
−
⎟ = x −
3!
5!
⎠
⎛
x2
x4
+
−
42. x cos x = x⎜1 −
2!
4!
⎝
⎞
x3
x5
+
−
⎟ = x−
2!
4!
⎠
43.
x2
x3
+
+
2!
3!
2 n +1
2
(
ex = 1 + x +
38.
(−1) ( x3 )
2∑
(2n + 1)!
n=0
∞
2 x3
2 x5
2 x7
+
+
+
3!
5!
7!
2n +1
2 x9
2 x15
= 2x −
+
−
3!
5!
ln x +
x2
x3
x4
x5
−
+
−
+
2!
3!
4!
5!
1 x
(e − e − x )
2
x3
x5
x7
= x +
+
+
+
3!
5!
7!
3
=
e− x = 1 − x +
367
sinh ( x) =
⎛
x9
x15
= 2⎜ x3 −
+
−
3!
5!
⎝
39. cos 2 ( x) =
x2
x3
x4
x5
+
+
+
+
2!
3!
4!
5!
x3
x6
+
−
2!
4!
(−1)
∞
ex = 1 + x +
e x − e− x = 2 x +
n 3n
∑ (2n)!
n=0
=1−
36.
x2
x4
+
−
2!
4!
(2n)!
n=0
=
=1−
Taylor and Maclaurin Series
x − ( x3 3!) + ( x5 5!) −
sin x
=
x
x
=1−
x2
x4
+
−
2!
4!
=
=
.
(−1)
∑
n = 0 ( 2n
n
∞
x2n + 2
+ 1)!
(−1) x 2 n +1
∑ (2n)!
n=0
n
∞
=
∞
(−1)
n
x2n
∑ (2n + 1)!, x
≠ 0
n=0
= 1, x = 0
© 2010 Brooks/Cole, Cengage Learning
368
44.
Chapter 9
arcsin x
=
x
Infinite Series
(2n)!x 2 n +1
2
n = 0 ( 2 n n!) ( 2n + 1)
∞
∑
1
=
x
⋅
(2n)!x 2 n , x
2
n = 0 ( 2 n n!) ( 2 n + 1)
∞
∑
≠ 0
= 1, x = 0
45.
(ix)2
eix = 1 + ix +
2!
(−ix)
e − ix = 1 − ix +
eix − e − ix
+
(ix)3
3!
2
+
2!
(ix)4
+
(−ix)
4!
3
+
3!
+
x2
ix3
x4
ix5
x6
−
+
+
−
−
2!
3!
4!
5!
6!
= 1 + ix −
(−ix)
4
+
4!
= 1 − ix −
x2
ix3
x4
ix5
x6
+
+
−
−
+
2!
3!
4!
5!
6!
2ix3
2ix5
2ix 7
= 2ix −
+
−
+
3!
5!
7!
eix − e − ix
x3
x5
x7
= x −
+
−
+
2i
3!
5!
7!
=
(−1)n x 2n +1
∑
n = 0 ( 2 n + 1)!
∞
2x2
2 x4
2 x6
+
−
+
2!
4!
6!
46. eix + e − ix = 2 −
eix + e − ix
x2
x4
x6
=1−
+
−
+
2
2!
4!
6!
= sin ( x)
(See Exercise 45.)
=
∞
(−1)n x 2n
∑ (2n)!
n=0
= cos( x)
47. f ( x) = e x sin x
14
⎛
x
x
x
= ⎜1 + x +
+
+
+
2
6
24
⎝
2
3
⎞⎛
x
x
+
−
⎟⎜ x −
6
120
⎠⎝
4
3
⎞
⎟
⎠
5
P5
f
⎛ x3
x3 ⎞ ⎛ x 4
x 4 ⎞ ⎛ x5
x5
x5 ⎞
= x+ x +⎜
−
−
−
+
⎟+⎜
⎟+⎜
⎟+
6⎠ ⎝6
6 ⎠ ⎝ 120 12
24 ⎠
⎝2
−6
2
= x + x2 +
P5 ( x) = x + x 2 +
6
−2
x3
x5
−
+
3
30
x3
x5
−
3
30
48. g ( x) = e x cos x
8
⎛
x2
x4
x4
= ⎜1 + x +
+
+
+
2
6
24
⎝
⎞⎛
x2
x4
+
−
⎟⎜1 −
2
24
⎠⎝
⎞
⎟
⎠
−6
6
g
P4
⎛ x2
x 2 ⎞ ⎛ x3
x3 ⎞ ⎛ x 4
x4
x4 ⎞
=1+ x +⎜
−
−
−
+
⎟+⎜
⎟+⎜
⎟+
2⎠ ⎝6
2 ⎠ ⎝ 24
4
24 ⎠
⎝2
−40
x3
x4
=1+ x −
−
+
3
6
P4 ( x) = 1 + x −
x3
x4
−
3
6
49. h( x) = cos x ln (1 + x)
⎛
x
x
= ⎜1 −
+
+
2
24
⎝
2
4
4
⎞⎛
x
x
x
x
+
−
+
−
⎟⎜ x −
2
3
4
5
⎠⎝
2
3
4
5
⎞
⎟
⎠
P5
−3
= x−
⎛x
x
x ⎞ ⎛x
x ⎞ ⎛x
x
x ⎞
+⎜
−
−
−
+
⎟+⎜
⎟+⎜
⎟+
2
3
2
4
4
5
6
24
⎝
⎠ ⎝
⎠ ⎝
⎠
= x−
x2
x3
3x5
−
+
+
2
6
40
P5 ( x) = x −
2
3
3
4
4
5
5
9
h
5
−4
x2
x3
3x5
−
+
2
6
40
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
Taylor and Maclaurin Series
369
50. f ( x) = e x ln (1 + x)
⎛
x2
x3
x4
= ⎜1 + x +
+
+
+
2
6
24
⎝
⎞⎛
x2
x3
x4
x5
+
−
+
−
⎟⎜ x −
2
3
4
5
⎠⎝
⎞
⎟
⎠
⎛
x 2 ⎞ ⎛ x3
x3
x3 ⎞ ⎛ x 4
x4
x4
x 4 ⎞ ⎛ x5
x5
x5
x5
x5 ⎞
= x + ⎜ x2 −
−
+
+
−
+
−
+
−
+
⎟ +⎜
⎟ + ⎜−
⎟+⎜
⎟+
2⎠ ⎝3
2
2⎠ ⎝ 4
3
4
6⎠ ⎝5
4
6
12
24 ⎠
⎝
x2
x3
3x5
+
+
+
2
3
40
= x +
x2
x3
3x5
+
+
2
3
40
P5 ( x) = x +
3
f
P5
−3
3
−3
51. g ( x) =
sin x
. Divide the series for sin x by (1 + x).
1+ x
52. f ( x) =
5 x3
5x4
−
+
6
6
x3
x5
1 + x x + 0 x2 −
+ 0 x4 +
+
6
120
x + x2
x3
− x2 −
6
− x 2 − x3
5 x3
+ 0x4
6
5 x3
5x4
+
6
6
5x4
x5
−
+
6
120
5x4
5 x5
−
−
6
6
3x 4
x2
x3
−
+
+
2
3
8
2
3
x
x
x4
x5
1+ x1+ x +
+
+
+
+
2
6
24 120
1+ x
x2
x3
0+
+
2
6
x2
x3
+
2
2
x3
x4
−
+
3
24
x3
x4
−
−
3
3
3x 4
x5
+
8
120
3x 4
3x5
+
8
8
x − x2 +
g ( x) = x − x 2 +
5 x3
5x4
−
+
6
6
P4 ( x) = x − x 2 +
5 x3
5x4
−
6
6
ex
. Divide the series for e x by (1 + x).
1+ x
1+
f ( x) = 1 +
3x 4
x2
x3
−
+
−
2
3
8
P4 ( x) = 1 +
3x 4
x2
x3
−
+
2
3
8
4
6
g
P4
−6
f
6
P4
−6
6
−4
−6
53.
y = x2 −
⎛
x4
x3 ⎞
= x⎜ x −
⎟
3!
3! ⎠
⎝
f ( x) = x sin x
Matches (c)
© 2010 Brooks/Cole, Cengage Learning
370
Chapter 9
y = x −
54.
Infinite Series
⎛
x3
x5
x2
x4 ⎞
+
= x⎜1 −
+
⎟
2!
4!
2!
4! ⎠
⎝
∫0 (
x
57.
)
2
x
e − t − 1 dt =
∫0
=
x
f ( x) = x cos x
Matches (d)
⎛
x3
x2 ⎞
= x⎜1 + x +
⎟
2!
2! ⎠
⎝
y = x + x2 +
55.
∫0
x
(−1)n +1 x 2 n + 3
∞
∑ (2n + 3)(n + 1)!
=
Matches (a)
n=0
y = x 2 − x3 + x 4 = x 2 (1 − x + x 2 )
f ( x) = x 2
⎡ ∞ ( −1)n +1 t 2 n + 2 ⎤
⎢∑
⎥ dt
⎢⎣n = 0 ( n + 1)! ⎥⎦
⎡ ∞ ( −1)n +1 t 2 n + 3 ⎤
= ⎢∑
⎥
⎢⎣n = 0 ( 2n + 3)( n + 1)!⎥⎦ 0
f ( x) = xe x
56.
⎡⎛ ∞ ( −1)n t 2 n ⎞
⎤
⎢⎜ ∑
⎟ − 1⎥ dt
⎜
⎟
n! ⎠
⎢⎣⎝ n = 0
⎥⎦
1
1+ x
Matches (b)
58.
x
∫0
x
∫0
1 + t 3 dt =
⎡
t3
⎢1 +
+
2
⎢⎣
59. Because ln x =
60. Because sin ( x) =
sin (1) =
∞
∞
∑ (2n + 1)!
∞
∑
n=0
(−1)
∞
∑
n=0
and
n
x 2 n +1
=1−
1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ( 2n − 3) x3n + 1
(3n
+ 1)2n n!
=
∞
2
+
n +1 1
n
(x
− 1)
3
3
≈ 0.6931.
x3
x5
x7
+
−
+
3!
5!
7!
1
1
1
+
−
+
3! 5! 7!
=
− 1)
2
∑ (−1)
n =1
= x −
22
23
+
+
2!
3!
(x
= ( x − 1) −
xn
x2
x3
=1+ x +
+
+
n!
2!
3!
you have e 2 = 1 + 2 +
62. Because e x =
n +1
1 1 1
+ − +
2 3 4
n=0
61. Because e x =
− 1)
x
− 3)t 3n + 1 ⎤
⎥
⎥⎦
0
n −1
n=2
∑
n = 0 ( 2n + 1)!
(−1)n
(−1)
n +1
n=0
you have ln 2 = 1 −
∞
∑
− 3)t 3n ⎤
⎥ dt
⎥⎦
2n n!
n=2
∞
x4
+
8
(−1)n ( x
∞
∑
(−1)n −11 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n
(−1)n −11 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n
∑
(3n + 1)2n n!
n=2
⎡
t4
= ⎢t +
+
8
⎢⎣
= x +
∞
∑
−
(x
− 1)
4
4
+
,
(0
< x ≤ 2)
(10,001 terms)
, you have
≈ 0.8415.
(4 terms)
,
∞
2n
≈ 7.3891.
n = 0 n!
∑
xn
x2
x3
x4
x5
=1+ x +
+
+
+
+
n!
2!
3!
4!
5!
1
1
1
1
1
e −1
= 1 − e −1 = 1 −
+
−
+
−
+
2! 3! 4! 5! 7!
e
(12 terms)
, you have e −1 = 1 − 1 +
=
∞
(−1)n −1
n =1
n!
∑
≈ 0.6321.
1
1
1
1
−
+
−
+
2! 3! 4! 5!
(6 terms)
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
Taylor and Maclaurin Series
371
63. Because
cos x =
(−1)n x 2 n
∑ (2n)!
n=0
∞
x2
x4
x6
x8
+
−
+
−
2!
4!
6!
8!
=1−
x2
x4
x6
x8
−
+
−
+
2!
4!
6!
8!
1 − cos x =
1 − cos
x
x3
x5
x7
=
−
+
−
+
x
2!
4!
6!
8!
=
(−1)n x 2 n + 2
∞
∑
n = 0 ( 2 n + 2)!
=
(−1)n x 2n +1
∑
n = 0 ( 2 n + 2)!
∞
∞
(−1) x 2 n +1 = 0.
1 − cos x
= lim ∑
x→0
x→0
x
n = 0 ( 2n + 2)!
you have lim
64. Because
sin x =
∞
(−1)
n
x
2 n +1
∑
n = 0 ( 2n + 1)!
sin x
x2
x4
x6
=1−
+
−
+
x
3!
5!
7!
=
∫0 e
67.
x3
x5
x7
+
−
+
3!
5!
7!
= x −
1
− x3
⎡ ∞ − x3 n ⎤
( ) ⎥ dx
dx = ∫ ⎢ ∑
0⎢
n! ⎥
n=0
⎣
⎦
1
(−1) x 2 n
∑ (2n + 1)!
n=0
n
∞
=
x→0
2
2
∞
x
x
+
+
2!
3!
ex − 1 = x +
and
3
3
=
n!
(−1)
∞
⎥⎦
0
n
∑ (3n + 1)n!
n=0
x
n +1
∑ (n + 1)!
=1−
n=0
∞
ex − 1
x
x2
=1+
+
+
2!
3!
x
(−1)n x3n ⎤⎥ dx
⎡ ∞ ( −1)n x3n + 1 ⎤
⎥
= ⎢∑
3n + 1)n! ⎥
⎣⎢n = 0 (
⎦
x
x
+
+
2!
3!
65. Because e x = 1 + x +
∫ 0 ⎢⎢n∑= 0
⎣
1
∞
(−1) x 2n = 1.
sin x
= lim ∑
x→0
x
n = 0 ( 2 n + 1)!
n
you have lim
1⎡ ∞
xn
∑ (n + 1)!
Because
n=0
ex − 1
xn
you have lim
= lim ∑
= 1.
x→0
x→0
x
(n + 1)!
1
∫0 e
− x2
1
1
+
−
4 14
+ ( −1)
n
(3n
1
+
+ 1)n!
1
< 0.0001, you need 6 terms.
⎣⎡3(6) + 1⎤⎦ 6!
dx ≈
5
(−1)n
∑ (3n + 1)n!
≈ 0.8075
n=0
x2
x3
+
−
2
3
66. Because ln ( x + 1) = x −
(See Exercise 29.)
ln ( x + 1)
x
=1−
you have lim
ln ( x + 1)
x→0
68.
14
∫0
69.
x
x ln ( x + 1) dx =
Because
(1 4)5
15
∞
x
x2
+
−
2
3
∑
n=0
= lim
∞
x→0
∑
n=0
(−1)n x n
n +1
(−1)n x n
n +1
= 1.
1 4⎛
∫0
⎞
⎡ x3
x4
x5
x6
+
−
+
⎟ dx = ⎢ −
4⋅2 5⋅3 6⋅4
⎠
⎣3
x3
x4
x5
2
+
−
+
⎜x −
2
3
4
⎝
< 0.0001, ∫
14
0
x ln ( x + 1) dx ≈
(1 4)3
3
−
(1 4)4
8
1
1⎡ ∞
≈ 0.00472.
n
n
(−1) x 2 n ⎤⎥ dx = ⎡⎢ ∞
(−1) x 2n +1 ⎤⎥ =
⎢
∑
∫ 0 ⎢n∑= 0 (2n + 1)!⎥
⎢⎣n = 0 ( 2n + 1)( 2n + 1)!⎥⎦ 0
⎣
⎦
Because 1 (7 ⋅ 7!) < 0.0001, you need three terms:
sin x
∫ 0 x dx =
1
sin x
1
1
dx = 1 −
+
−
x
3 ⋅ 3! 5 ⋅ 5!
sin x
Note: You are using lim
= 1.
x
x → 0+
1
∫0
≈ 0.9461.
14
⎤
⎥
⎦0
∞
∑ ( 2n
n=0
(−1)
+ 1)( 2n
n
+ 1)!
(using three nonzero terms)
© 2010 Brooks/Cole, Cengage Learning
372
70.
Chapter 9
1
∫0
Infinite Series
cos x 2 dx =
(−1)n x 4n ⎤⎥ dx
(2n)! ⎥⎦
1⎡ ∞
∫ 0 ⎢⎢n∑= 0
⎣
72.
12
∫0
arctan ( x 2 ) dx =
(−1)
n
(−1)
n
∑ (4n + 1)(2n)!
1
Because
71.
12
∫0
2
dx ≈
3
∑ (4n + 1)(2n)!
Because
≈ 0.904523
(−1)
n
1 2⎛
1
< 0.0001
+ 3)( 2n + 1)24 n + 3
12
∫0
1
1
−
≈ 0.041295
3(1) ⋅ 23
7(3)27
arctan( x 2 ) dx ≈
⎞
⎟ dx
⎠
x2
x4
x6
+
−
+
⎜1 −
3
5
7
⎝
⎡
x3
x5
x7
= ⎢x − 2 + 2 − 2 +
3
5
7
⎣
( 4n
when n = 2, you need 2 terms.
1
< 0.0001, you need 4 terms.
⎡⎣4( 4) + 1⎤⎡
⎦⎣2( 4)⎤⎦!
∫0
∞
n=0
n=0
arctan x
dx =
x
2n + 1 ⎥⎦
∑ (4n + 3)(2n + 1)24n + 3
=
n=0
∫ 0 cos x
(−1)n x 4n + 2 ⎤⎥ dx
⎡∞
(−1)n x 4 n + 3 ⎤⎥
= ⎢∑
⎢⎣n = 0 ( 4n + 3)( 2n + 1) ⎥⎦
0
⎡ ∞ ( −1)n x 4 n + 1 ⎤
⎥
= ⎢∑
⎢⎣n = 0 ( 4n + 1)( 2n)!⎥⎦
0
∞
∫ 0 ⎢⎢n∑= 0
⎣
12
1
=
1⎡ ∞
12
⎤
⎥
⎦0
Because 1 (92 29 ) < 0.0001, you have
12
∫0
arctan x
1
1
1
1 ⎞
⎛1
dx ≈ ⎜ − 2 3 + 2 5 − 2 7 + 2 9 ⎟
x
52
72
92 ⎠
⎝2 3 2
≈ 0.4872.
Note: You are using lim
x → 0+
73.
0.3
∫ 0.1
1 + x3 dx =
Because
0.3
∫ 0.1
∫0
2
= (1 + x
1 + x 2 dx =
Because
0.2
∫ 0.1
x3
x6
x9
5 x12
−
+
−
+
⎜1 +
2
8
16
128
⎝
⎞
⎡
x4
x7
x10
5 x13
−
+
−
+
⎟ dx = ⎢ x +
8
56 160 1664
⎠
⎣
0.3
⎤
⎥
⎦ 0.1
1
(0.37 − 0.17 ) < 0.0001, you need two terms.
56
1+ x
∫0
0.3 ⎛
1
⎡
⎤
1 + x3 dx = ⎢(0.3 − 0.1) + (0.34 − 0.14 )⎥ ≈ 0.201.
8
⎣
⎦
74.
0.2
arctan x
= 1.
x
(0.2)5
40
0.2
∫0
2
)
12
1⎛ 1 ⎞ 4
1 ⎛ 1 ⎞⎛ 3 ⎞ 6
⎜− ⎟x
⎜ − ⎟⎜ − ⎟ x
1 2
2⎝ 2 ⎠
2 2
2
=1+ x +
+ ⎝ ⎠⎝ ⎠ +
2
2!
3!
1 2 1 4
1 6
⎡
⎢1 + 2 x − 8 x + 16 x −
⎣
= 1+
⎡
x3
x5
x7
⎤
⎥ dx = ⎢ x + 6 − 40 + 112 −
⎦
⎣
1 2 1 4
1 6
x − x +
x −
2
8
16
0.2
⎤
⎥
⎦0
< 0.0001, you need 2 terms.
1 + x 2 dx ≈ 0.2 +
(0.2)3
6
≈ 0.201333
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
Taylor and Maclaurin Series
373
π 2
75.
π 2
∫0
∫0
x cos x dx =
Because 2(π 2)
⎡
⎢∑
⎢⎣n = 0
(23 ⋅ 10!)
23 2
⎡
⎤
⎢ ∞ ( −1)n x(4 n + 3) 2 ⎥
⎥
⎥ dx = ⎢ ∑
⎢n = 0 ⎛ 4n + 3 ⎞ 2n !⎥
⎥⎦
(
)
⎜
⎟
⎢ ⎝ 2 ⎠
⎥
⎣
⎦0
π 2
(−1) x(4n +1) 2 ⎤
n
(2n)!
⎡ ∞ ( −1) n 2 x(4 n + 3) 2 ⎤
⎥
= ⎢∑
⎢⎣n = 0 ( 4n + 3)( 2n)! ⎥⎦
0
< 0.0001, you need five terms.
72
11 2
15 2
19 2
⎡ (π 2)3 2
(π 2) + (π 2) − (π 2) + (π 2) ⎤⎥ ≈ 0.7040.
x cos x dx = 2 ⎢
−
14
264
10,800
766,080 ⎥
⎢⎣ 3
⎦
1
∫0
76.
π 2
∞
⎡
⎞
x2
x3
x4
x5
+
−
+
−
⎟ dx = ⎢ x −
2( 2!)
3( 4!)
4(6!)
5(8!)
⎢⎣
⎠
⎛
x
x2
x3
x4
+
−
+
−
x dx = ∫ ⎜1 −
0.5
2!
4!
6!
8!
⎝
1
1
∫ 0.5 cos
1
⎤
⎥
⎦⎥ 0.5
1
(1 − 0.55 ) < 0.0001, you have
201,600
Because
1
1
1
1
⎡
x dx ≈ ⎢(1 − 0.5) − (1 − 0.52 ) +
(1 − 0.53 ) − 2880
(1 − 0.54 ) + 201,600
(1 − 0.55 )⎤⎥ ≈ 0.3243.
4
72
⎣
⎦
1
∫ 0.5 cos
77. From Exercise 27, you have
1
2π
1
∫0
e− x
2 2
1 ∞
(−1)
n
x2n
1
1
2π
∫ 0 n∑= 0
≈
1
2π
⎛
⎞
1
1
1
+ 2
− 3
⎜1 −
⎟ ≈ 0.3412.
2
1
3
2
2!
5
2
3!
7
⋅
⋅
⋅
⋅
⋅
⋅
⎝
⎠
2n n!
dx =
1
2π
⎡ ∞ ( −1) n x 2 n + 1 ⎤
⎢∑ n
⎥ =
⎢⎣n = 0 2 n!( 2n + 1) ⎥⎦ 0
dx =
1
2π
(−1)
∑ 2n n!(2n + 1)
n=0
n
∞
78. From Exercise 27, you have
1
2π
2
∫1
e− x
2 2
dx =
1
2π
=
1
2π
≈
79. f ( x) = x cos 2 x =
P5 ( x) = x − 2 x3 +
2 ∞
∫ 1 n∑= 0
∞
∑
n=0
(−1)
n
x 2n
2n n!
(−1)
n
2
dx =
n
1 ⎡ ∞ ( −1) x 2 n + 1 ⎤
⎢∑ n
⎥
2π ⎣⎢n = 0 2 n!( 2n + 1) ⎦⎥
1
(2n +1 − 1)
2 n!( 2n + 1)
n
1 ⎛
7
31
127
511
2047 ⎞
+ 2
− 3
+ 4
− 5
⎜1 −
⎟
2 ⋅ 1 ⋅ 3 2 ⋅ 2! ⋅ 5 2 ⋅ 3! ⋅ 7
2 ⋅ 4! ⋅ 9
2 ⋅ 5! ⋅ 11 ⎠
2π ⎝
8191
32,767
131,071
524,287
− 9
≈ 0.1359.
+ 6
− 7
+ 8
2 ⋅ 6! ⋅ 13 2 ⋅ 7! ⋅ 15
2 ⋅ 8! ⋅ 17
2 ⋅ 9! ⋅ 19
(−1) 4n x 2 n +1
(2n)!
n=0
∞
∑
n
2 x5
3
x
ln (1 + x)
2
7 x4
11x5
x2
x3
P5 ( x) =
−
+
−
2
4
48
96
80. f ( x) = sin
4
2
f
f
−4
−3
8
3
P5
P5
−4
−2
The polynomial is a reasonable approximation on the
⎡ 3 3⎤
interval ⎢− , ⎥.
⎣ 4 4⎦
The polynomial is a reasonable approximation on the
interval ( −0.60, 0.73).
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
374
81. f ( x) =
Infinite Series
x ln x, c = 1
P5 ( x) = ( x − 1) −
(x
− 1)
24
3
+
(x
− 1)
24
4
−
71( x − 1)
5
1920
⎡1 ⎤
The polynomial is a reasonable approximation on the interval ⎢ , 2⎥.
⎣4 ⎦
3
P5
g
−2
4
−2
82. f ( x) =
3
x ⋅ arctan x, c = 1
⎡ ( x − 1)2 ⎤
⎡ ( x − 1)3 ⎤
⎡ ( x − 1)4 ⎤
⎡ ( x − 1)5 ⎤
P5 ( x) ≈ 0.7854 + 0.7618( x − 1) − 0.3412 ⎢
⎥ − 0.0424 ⎢
⎥ + 1.3025 ⎢
⎥ − 5.5913 ⎢
⎥
⎢⎣ 2! ⎥⎦
⎢⎣ 3! ⎥⎦
⎢⎣ 4! ⎥⎦
⎢⎣ 5! ⎥⎦
The polynomial is a reasonable approximation on the interval (0.48, 1.75).
3
f
P5
−2
4
−1
83. See Guidelines, page 682.
84. a2 n + 1 = 0 (odd coefficients are zero)
85. The binomial series is (1 + x) = 1 + kx +
k
k ( k − 1)
2!
x2 +
k ( k − 1)( k − 2)
3!
x3 +
. The radius of convergence is R = 1.
86. (a) Replace x with ( − x).
(b) Replace x with 3 x.
(c) Multiply series by x.
(d) Replace x with 2 x, then replace x with −2 x, and add the two together.
⎛
⎞
g
g ⎛
kx ⎞
87. y = ⎜ tan θ −
⎟ x − 2 ln ⎜1 −
⎟
kv0 cos θ ⎠
k
v0 cos θ ⎠
⎝
⎝
= ( tan θ ) x −
2
3
4
gx
g⎡
kx
1 ⎛ kx ⎞
1 ⎛ kx ⎞
1 ⎛ kx ⎞
− 2 ⎢−
− ⎜
⎟ − ⎜
⎟ − ⎜
⎟ −
kv0 cos θ
k ⎢ v0 cos θ
2 ⎝ v0 cos θ ⎠
3 ⎝ v0 cos θ ⎠
4 ⎝ v0 cos θ ⎠
⎣
= ( tan θ ) x −
gx
gx
gx 2
gkx3
gk 2 x 4
+
+
+
+
+
2
2
3
3
kv0 cos θ
kv0 cos θ
2v0 cos θ
3v0 cos θ
4v0 4 cos 4 θ
= ( tan θ ) x +
gx 2
kgx3
k 2 gx 4
+
+
+
2
3
3
2v0 cos θ
3v0 cos θ
4v0 4 cos 4 θ
⎤
⎥
⎥⎦
2
© 2010 Brooks/Cole, Cengage Learning
Section 9.10
88. θ = 60°, v0 = 64, k =
y =
2(64) (1 2)
2
−
2
(1 16)(32) x3
3
3
3(64) (1 2)
−
(1 16)2 (32) x 4
4
4
4(64) (1 2)
=
⎡ 22 x 2
23 x 3
24 x 4
3x − 32 ⎢
+
+
+
2
3
4
2
3(64) 16
4(64) (16)
⎣⎢ 2(64)
=
3x − 32 ∑
∞
2n x n
n=2
⎪⎧e −1
89. f ( x) = ⎨
⎪⎩0,
x2
375
1
, g = −32
16
32 x 2
3x −
Taylor and Maclaurin Series
n(64) (16)
n
∞
3x − 32 ∑
=
n−2
n=2
−
⎤
⎥
⎦⎥
xn
n(32) (16)
n
n−2
, x ≠ 0
x = 0
(a)
y
2
1
x
−3 −2 −1
1
(b) f ′(0) = lim
2
3
f ( x ) − f ( 0)
x −0
x→0
Let y = lim
e
= lim
e −1
x2
−0
x
x→0
−1 x 2
. Then
x
x→0
⎛ e −1 x2
ln y = lim ln ⎜
x→0
⎜ x
⎝
⎞
⎡ −1 − x 2 ln x ⎤
1
⎟ = lim ⎡⎢− 2 − ln x⎤⎥ = lim ⎢
⎥ = −∞.
+
+
⎟
x2
x→0 ⎣ x
x→0 ⎣
⎦
⎦
⎠
So, y = e −∞ = 0 and you have f ′(0) = 0.
(c)
∞
f (n) (0)
n=0
n!
∑
90. (a) f ( x) =
x n = f (0) +
ln ( x 2 + 1)
x2
f ′(0) x
1!
+
f ′′(0) x 2
2!
+
= 0 ≠ f ( x) This series converges to f at x = 0 only.
.
From Exercise 10, you obtain:
1 ∞ ( −1) x 2 n + 2
=
∑
x2 n = 0 n + 1
n
P =
P8 = 1 −
(b)
2
4
6
(−1)n x 2 n
∞
∑
n=0
n +1
8
x
x
x
x
+
−
+ .
2
3
4
5
1.5
0
2
−0.5
© 2010 Brooks/Cole, Cengage Learning
376
Chapter 9
(c) F ( x) =
G ( x) =
Infinite Series
x
∫0
ln (t 2 + 1)
t2
dt
x
∫ 0 P8 (t ) dt
x
0.25
0.50
0.75
1.00
1.50
2.00
F ( x)
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
G ( x)
0.2475
0.4810
0.6924
0.8865
1.6878
9.6063
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval.
91. By the Ratio Test: lim
n→∞
x
x n +1
n!
⋅ n = lim
= 0 which shows that
n
→
∞
n +1
(n + 1)! x
∞
∑
n=0
xn
converges for all x.
n!
⎛1 + x ⎞
92. ln ⎜
⎟ = ln (1 + x) − ln (1 − x)
⎝1 − x ⎠
⎛
x2
x3
= ⎜x −
+
−
2
3
⎝
⎞ ⎛
x2
x3
−
−
⎟ − ⎜−x −
2
3
⎠ ⎝
∞
⎞
x3
x5
⎟ = 2x + 2 + 2 +
3
5
⎠
x 2n
,R = 1
n = 0 2n + 1
) = 2x∑
⎛1 + 1 2 ⎞
(1 2)2 + (1 2)4 + (1 2)6 ⎤⎥ = 1 + 1 + 1 + 1 ≈ 1.098065
⎛1⎞⎡
⎢
ln 3 = ln ⎜
2
1
≈
+
⎟
⎜ ⎟
3
5
7 ⎥
12 80
448
⎝ 2 ⎠ ⎢⎣
⎝1 − 1 2 ⎠
⎦
(ln 3
≈ 1.098612)
⎛ −1 3⎞
(−1 3)(−4 3)(−7 3)(−10 3)(−13 3)
96. ⎜
⎟ =
5!
5
⎝
⎠
−91
=
≈ −0.12483
729
⎛ 5⎞
5⋅4⋅3
60
=
= 10
93. ⎜ ⎟ =
3!
6
3
⎝ ⎠
⎛ −2 ⎞
(−2)(−3) = 3
94. ⎜ ⎟ =
2!
2
⎝ ⎠
97. (1 + x) =
k
⎛ 0.5 ⎞ (0.5)( −0.5)( −1.5)( −2.5)
5
95. ⎜ ⎟ =
= −0.0390625 = −
4!
128
⎝ 4 ⎠
∞
⎛k ⎞
∑ ⎜ n ⎟x n
n=0 ⎝
⎠
Example: (1 + x) =
2
∞
⎛ 2⎞
∑ ⎜ n ⎟x n
n=0 ⎝
⎠
= 1 + 2x + x2
98. Assume e = p q is rational. Let N > q and form the following.
1
⎡
e − ⎢1 + 1 +
+
2!
⎣
+
1⎤
1
1
=
+
+
N !⎥⎦
( N + 1)! ( N + 2)!
⎡
⎛
Set a = N !⎢e − ⎜1 + 1 +
⎝
⎣
+
1 ⎞⎤
⎟ , a positive integer. But,
N !⎠⎥⎦
⎡ 1
1
a = N !⎢
+
+
N
N
+
+ 2)!
1
!
(
)
(
⎣⎢
=
1
N +
⎤
1
1
+
+
⎥ =
N
N
N + 2)
+
+
1
1
(
)(
⎦⎥
⎡
⎤
1
1
1
⎢1 +
+
+ …⎥ =
2
1 ⎢⎣
N + 1 ( N + 1)
N +
⎥⎦
<
1
1
+
+
N + 1 ( N + 1)2
1
1
⎡
⎤
= , a contradiction.
1⎢
⎛ 1 ⎞⎥
N
⎢1 − ⎜
⎟⎥
⎢⎣
⎝ N + 1 ⎠ ⎥⎦
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
99. g ( x) =
377
x
= a0 + a1 x + a2 x 2 +
1 − x − x2
x = (1 − x − x 2 )( a0 + a1 x + a2 x 2 +
)
x = a0 + ( a1 − a0 ) x + ( a2 − a1 − a0 ) x 2 + ( a3 − a2 − a1 ) x3 +
Equating coefficients,
a0 = 0
a1 − a0 = 1 ⇒ a1 = 1
a2 − a1 − a0 = 0 ⇒ a2 = 1
a3 − a2 − a1 = 0 ⇒ a3 = 2
a4 = a3 + a2 = 3, etc.
In general, an = an −1 + an − 2 . The coefficients are the Fibonacci numbers.
100. Assume the interval is [−1, 1]. Let x ∈ [−1, 1],
f (1) = f ( x) + (1 − x) f ′( x) +
1
2
f ( −1) = f ( x) + ( −1 − x) f ′( x) +
(1 − x)
1
2
2
f ′′(c), c ∈ ( x, 1)
(−1 − x)2 f ′′(d ), d ∈ (−1, x).
So, f (1) − f ( −1) = 2 f ′( x) +
1
2
(1 − x)
2
f ′′(c) −
1
2
(1 + x)
2
f ′′( d )
2 f ′( x ) = f (1) − f ( −1) −
1
2
(1 − x)
2
f ′′(c) +
1
2
(1 + x)
2
f ′′( d ).
Because f ( x) ≤ 1 and f ′′( x) ≤ 1,
2 f ′( x) ≤ f (1) + f ( −1) +
1
2
(1 − x)2
f ′′(c) +
1
2
(1 + x)2
f ′′( d ) ≤ 1 + 1 +
1
2
(1 − x2 ) + 12 (1 + x)2
= 3 + x 2 ≤ 4.
So, f ′( x) ≤ 2.
Note: Let f ( x) =
1
2
(x
+ 1) − 1. Then f ′( x) ≤ 1, f ′′( x) = 1 and f ′(1) = 2.
2
Review Exercises for Chapter 9
1. an =
( )
1
n! + 1
6. an = 6 − 23
n −1
: 6, − 4, …
Matches (b)
n
2. an = 2
n +1
3. an = 4 +
7. an =
2
: 6, 5, 4.67, …
n
Matches (a)
4. an = 4 −
The sequence seems to converge to 5.
n→∞
8
Matches (c)
Matches (d)
5n + 2
n
2⎞
⎛
= lim ⎜ 5 + ⎟ = 5
n→∞ ⎝
n⎠
lim an = lim
n→∞
n
: 3.5, 3, …
2
5. an = 10(0.3)
5n + 2
n
n −1
: 10, 3, …
0
12
0
© 2010 Brooks/Cole, Cengage Learning
Chapter 9
378
Infinite Series
nπ
2
8. an = sin
The sequence seems to diverge (oscillates).
sin
y = (b n + c n )
1n
18. Let
ln y =
nπ
: 1, 0, −1, 0, 1, 0, …
2
ln (b n + c n )
n
lim ln y = lim
n→∞
2
n→∞
1
(bn ln b + c n ln c).
b + cn
n
Assume b ≥ c and note that the terms
0
b n ln b + c n ln c
b n ln b
c n ln c
=
+
bn + cn
bn + cn
bn + cn
12
converge as n → ∞. Hence an converges.
−2
( 78 )
9. lim ⎡
⎢
n→∞ ⎣
n
n
+ 3⎤⎥ = 3
⎦
0.05 ⎞
⎛
19. (a) An = 8000⎜1 +
⎟ ,
4 ⎠
⎝
Converges
1
0.05 ⎞
⎛
A1 = 8000⎜1 +
⎟ = $8100.00
4 ⎠
⎝
A2 = $8201.25
5 ⎤
⎡
=1
10. lim ⎢1 +
n→∞ ⎣
n + 1⎥⎦
A3 = $8303.77
Converges
A4 = $8407.56
n3 + 1
= ∞
11. lim
n→∞
n2
A5 = $8512.66
A6 = $8619.07
Diverges
A7 = $8726.80
A8 = $8835.89
1
= 0
n
12. lim
n→∞
(b) A40 = $13,148.96
Converges
20. (a) Vn = 175,000(0.70) ,
n
n
13. lim 2
= 0
n→∞ n + 1
5
21. (a)
n
1
14. lim
= lim
= ∞
n →∞ ln ( n)
n →∞ 1 n
(
n +1 −
)
n = lim
n →∞
n →∞
(
n +1 −
n
1
n +1 +
n
)
n +1 +
n +1 +
n
n
(b)
10
15
20
25
Sn
13.2
113.3
873.8
6648.5
50,500.3
120
0
22. (a)
12
= e1 2
Converges; k = 2n
n
n
)
> 1.
12
−10
n
k
⎡⎛
1 ⎞
1⎞ ⎤
⎛
=
+
16. lim ⎜1 +
lim
1
⎢
⎟
⎜
⎟ ⎥
n→∞ ⎝
k →∞ ⎝
k ⎠ ⎥⎦
2n ⎠
⎢⎣
sin
3
2
=0
Converges
n→∞
5
The series diverges geometric r =
= lim
17. lim
n
(
Diverges
n →∞
n = 1, 2, …
(b) V5 = 175,000(0.70) ≈ $29,412.25
Converges
15. lim
n = 1, 2, 3, …
n
5
10
15
20
25
Sn
0.3917
0.3228
0.3627
0.3344
0.3564
The series converges by the Alternating Series Test.
(b)
1
= 0
Converges
0
12
0
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
23. (a)
n
5
10
15
20
25
Sn
0.4597
0.4597
0.4597
0.4597
0.4597
379
The series converges by the Alternating Series Test.
(b)
1
0
12
− 0.1
24. (a)
n
5
10
15
20
25
Sn
0.8333
0.9091
0.9375
0.9524
0.9615
The series converges, by the Limit Comparison Test with
(b)
1
∑ n2 .
1
0
12
0
∞
25.
n
2
⎛ 2⎞
∑ ⎜ ⎟ Geometric series with a = 1 and r = 3.
n=0 ⎝ 3 ⎠
S =
a
1
1
=
=
= 3
1− r
1 − ( 2 3)
13
n
∞
2n + 2
⎛ 2⎞
= 4 ∑ ⎜ ⎟ = 4(3) = 12
n
3
n=0
n=0 ⎝ 3 ⎠
∞
26.
∑
See Exercise 25.
∞
27.
∑ ⎡⎣(0.6)
n
n =1
n
+ (0.8) ⎤ =
⎦
∞
∑ 0.6(0.6)
n=0
⎡⎛ 2 ⎞ n
⎤
1
∑ ⎢⎜⎝ 3 ⎟⎠ − (n + 1)(n + 2) ⎥ =
⎥⎦
n=0 ⎢
⎣
∞
28.
=
∞
n
+
∞
∑ 0.8(0.8)
n
n
1
n=0
⎛ 2⎞
∑ ⎜⎝ 3 ⎟⎠ −
n=0
∞
⎛
1
1
6 10
8 10
11
= (0.6)
+ (0.8)
=
⋅
+
⋅
=
= 5.5
1 − 0.6
1 − 0.8
10 4
10 2
2
∑ ⎜⎝ n + 1 −
n=0
1 ⎞
⎟
n + 2⎠
⎡⎛
1
1⎞ ⎛ 1 1⎞ ⎛1 1⎞
− ⎜1 − ⎟ + ⎜ − ⎟ + ⎜ − ⎟ +
1 − ( 2 3) ⎢⎣⎝
2 ⎠ ⎝ 2 3⎠ ⎝ 3 4 ⎠
29. (a) 0.09 = 0.09 + 0.0009 + 0.000009 +
= 0.09(1 + 0.01 + 0.0001 +
)
=
⎤
⎥ = 3−1= 2
⎦
∞
∑ (0.09)(0.01)
n
n=0
(b) 0.09 =
0.09
1
=
1 − 0.01 11
30. (a) 0.64 = 0.64 + 0.0064 + 0.000064 +
= 0.64(1 + 0.01 + 0.0001 +
)
∞
= 0.64 ∑ (0.01)
n
n=0
(b) 0.64 =
0.64
64
=
1 − 0.01
99
31. Diverges. Geometric series with a = 1 and r = 1.67 > 1.
© 2010 Brooks/Cole, Cengage Learning
380
Chapter 9
Infinite Series
32. Converges. Geometric series with a = 1 and
r = 0.67 < 1.
42.
⎛1
34. Diverges. nth-Term Test, lim an = 23 .
n→∞
43.
35. D1 = 8
−
n =1
∞
1
∑ 2n
n =1
∞
6
∑ 5n − 1
⎡ 6 (5n − 1) ⎤
6
lim ⎢
⎥ =
n
1
5
n→∞⎣
⎦
D = 8 + 16(0.7) + 16(0.7) +
= −8 +
∞
∑ 16(0.7)
n
= −8 +
n=0
39
∑ 42,000(1.055)
By a limit comparison test with the divergent harmonic
∞
1
series ∑ , the series diverges.
n =1 n
+ 16(0.7) +
2
n
16
1
= 45 meters
1 − 0.7
3
44.
n
∞
n
∑
n + 3n
3
n =1
n=0
42,000(1 − 1.05540 )
1 − 1.055
lim
≈ $5,737,435.79
(
)
300 e0.06(2) − 1
e0.06 12 − 1
n3 2
n3 + 3n
=1
By a limit comparison test with the divergent p-series
∞
1
∑ n1 2 , the series diverges.
n =1
P(e rt − 1)
e r 12 − 1
n3 + 3n
= lim
1 n
n→∞
n
n→∞
37. (See Exercise 116 in Section 9.2)
=
1
n =1
D2 = 0.7(8) + 0.7(8) = 16(0.7)
A =
∞
∑ n2
Because the first series is a convergent p-series while the
second series is a convergent geometric series, the
difference converges.
n→∞
=
1⎞
⎟ =
2n ⎠
−
n =1
33. Diverges. nth-Term Test. lim an ≠ 0.
36. S =
∞
∑ ⎜⎝ n2
45.
≈ $7630.70
∞
1
∑
n3 + 2n
n =1
lim
38. (See Exercise 116 in Section 9.2)
n3 + 2n
1
1 ( n3 2 )
n→∞
12 t
⎤
r ⎞
⎛ 12 ⎞ ⎡⎛
A = P⎜ ⎟ ⎢⎜1 + ⎟ − 1⎥
12 ⎠
⎝ r ⎠ ⎢⎣⎝
⎥⎦
12(10)
⎤
0.035 ⎞
⎛ 12 ⎞ ⎡⎛
= 125⎜
− 1⎥ ≈ $17,929.06
⎟ ⎢⎜1 +
⎟
12 ⎠
⎝ 0.035 ⎠ ⎣⎢⎝
⎥⎦
= lim
n→∞
n3 2
n3 + 2n
=1
By a limit comparison test with the convergent p-series
∞
1
∑ n3 2 , the series converges.
n =1
∞
46.
n +1
∑ n(n + 2)
n =1
39.
∞
∫1
b
1 ⎤
1
1
⎡ ln x
=
x −4 ln ( x) dx = lim ⎢− 3 − 3 ⎥ = 0 +
b → ∞ ⎣ 3x
9 x ⎦1
9
9
lim
By the Integral Test, the series converges.
∞
40.
∑4
n =1
1
n
=
3
∞
∞
⎛1
∑ ⎜⎝ n2
1
n =1
∞
n +1
=1
+ 2
n→∞ n
1
∑ n , the series
diverges.
1
∑ n2
n =1
= lim
∞
∑ n3 4
1⎞
⎟ =
n⎠
1n
By a limit comparison test with
n =1
−
+ 1) n( n + 2)
n =1
Divergent p-series, p =
41.
(n
n→∞
3
<1
4
−
∞
1
∑n
n =1
Because the second series is a divergent p-series while
the first series is a convergent p-series, the difference
diverges.
47.
∞
∑
(2n − 1)
2 ⋅ 4 ⋅ 6 ( 2 n)
1⋅3⋅5
n =1
an =
1 ⋅ 3 ⋅ 5 ( 2n − 1) ⎛ 3 5
= ⎜ ⋅
2 ⋅ 4 ⋅ 6 ( 2 n)
⎝2 4
Because
∞
1
∑ 2n
n =1
=
2n − 1 ⎞ 1
1
>
⎟
2n − 2 ⎠ 2n
2n
1 ∞ 1
∑ diverges (harmonic series), so
2 n =1 n
does the original series.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
48. Because
∞
∞
1
∑ 3n converges, ∑ 3n
n =1
n =1
1
converges by the
−5
54. Converges by the Alternating Series Test.
an + 1 =
Limit Comparison Test.
∞
49.
∑
(−1)n
n5
n =1
lim
n→∞
∞
50.
∑
converges by the Alternating Series Test.
(−1) (n
+ 1)
n2 + 1
n =1
55. lim
56. lim
converges by the Alternating Series
51.
∑
∞
−( x + 2 x − 1)
x +1
, f ′( x) =
< 0 ⇒ terms
2
2
x +1
( x 2 + 1)
lim
n→∞
57.
∞
∑
(−1)
n =1
an + 1
lim
n→∞
⎛ 4n ⎞
⎜
⎟
⎝ 7n − 1 ⎠
n
4
⎛ 4n ⎞
= lim ⎜
<1
⎟ =
n → ∞ ⎝ 7n − 1 ⎠
7
2
an + 1
n + 1 en
= lim
⋅
n → ∞ ( n + 1)2
an
n
e
= lim
n→∞
e n ( n + 1)
−( n 2 + 3)
n
, f ′( x) =
< 0 ⇒ terms are
2
n −3
(n2 − 3)
n
en
2 + 2n +1
n
⎛ 1 ⎞⎛ n + 1 ⎞
= lim ⎜ 2 n + 1 ⎟⎜
⎟
n→∞ ⎝ e
⎠⎝ n ⎠
= (0)(1) = 0 < 1
2
By the Ratio Test, the series converges.
∞
58.
n!
∑ en
n =1
n
lim
n→∞
n +1
n
=
≤
= an
n + 2
n +1
(n + 1)! ⋅ en
an + 1
= lim
n→∞
an
en +1
n!
= lim
n→∞
n
= 0
n +1
n +1
= ∞
e
By the Ratio Test, the series diverges.
∞
59.
2n
∑ n3
n =1
53. Diverges by the nth-Term Test.
n→∞
3
⎛ 3n − 1 ⎞
= lim ⎜
>1
⎟ =
n → ∞ ⎝ 2n + 5 ⎠
2
2
converges by the Alternating Series Test.
By the Alternating Series Test, the series converges.
lim
en
lim
n +1
n =1
n
3 ln n
3 ln n
= an , lim
= 0
n→∞
n
n
n
n→∞
decreasing. So, an + 1 < an .
52.
∑
n
= 0 and if
n2 − 3
f ( x) =
⎛ 3n − 1 ⎞
⎜
⎟
⎝ 2n + 5 ⎠
<
2
− n
n2 − 3
n=2
n +1
Converges by Root Test.
2
(−1)n
n
n→∞
are decreasing. So, an + 1 < an .
∞
3 ln ( n + 1)
Diverges by Root Test.
n +1
Test. lim 2
= 0 and if
n→∞ n + 1
f ( x) =
n
n→∞
1
1
1
< 5 = an .
= 0 and an + 1 =
5
n5
n
(n + 1)
n
381
n
=1 ≠ 0
n −3
lim
n→∞
an + 1
2n +1
n3
2n 3
= lim
⋅ n = lim
= 2
3
n→∞ n + 1
n→∞ n + 1 3
an
(
) 2
(
)
By the Ratio Test, the series diverges.
60.
∞
1⋅3⋅5
( 2n
− 1)
∑ 2 ⋅ 5 ⋅ 8 (3n − 1)
n =1
lim
n→∞
1⋅3
an + 1
= lim
n→∞ 2 ⋅ 5
an
( 2n
(3n
− 1)( 2n + 1)
− 1)(3n + 2)
⋅
2⋅5
1⋅3
(3n − 1)
(2n − 1)
= lim
n→∞
2n + 1
2
=
3n + 2
3
By the Ratio Test, the series converges.
© 2010 Brooks/Cole, Cengage Learning
382
Chapter 9
Infinite Series
61. (a) Ratio Test: lim
n→∞
(b)
(c)
(n + 1)(3 5)
an + 1
= lim
n
n
→
∞
an
n(3 5)
n +1
3
⎛ n + 1 ⎞⎛ 3 ⎞
= lim ⎜
< 1, Converges
⎟⎜ ⎟ =
n→∞ ⎝
n ⎠⎝ 5 ⎠
5
n
5
10
15
20
25
Sn
2.8752
3.6366
3.7377
3.7488
3.7499
4
0
12
−1
(d) The sum is approximately 3.75.
62. (a) The series converges by the Alternating Series Test.
(b)
(c)
n
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
0.3
0
12
0
(d) The sum is approximately 0.0714.
63. (a)
∞
1
1
⎡ 1⎤
dx = ⎢− ⎥ =
x2
x
N
⎣ ⎦N
∞
∫N
N
N
1
∑ n2
5
10
20
30
40
1.4636
1.5498
1.5962
1.6122
1.6202
0.2000
0.1000
0.0500
0.0333
0.0250
n =1
∞
∫N
(b)
1
dx
x2
∞
1
1
⎡ 1 ⎤
dx = ⎢− 4 ⎥ =
x5
4N 4
⎣ 4x ⎦ N
∞
∫N
N
N
1
∑ n5
5
10
20
30
40
1.0367
1.0369
1.0369
1.0369
1.0369
0.0004
0.0000
0.0000
0.0000
0.0000
n =1
∞
∫N
1
dx
x5
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums.
64. No. Let an =
3937.5
, then a75 = 0.7. The series
n2
∞
3937.5
is a convergent p-series.
n2
n =1
∑
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
65.
f ( x ) = e −3 x
383
f (0) = 1
f ′( x) = −3e −3 x
f ′(0) = −3
f ′′( x) = 9e −3 x
f ′′(0) = 9
f ′′′( x) = −27e−3 x
f ′′′(0) = −27
9x2
27 x3
−
2!
3!
9 2
9 3
= 1 − 3x + x − x
2
2
P3 ( x) = 1 − 3x +
66.
f ( x) = tan x
⎛ π⎞
f ⎜ − ⎟ = −1
⎝ 4⎠
f ′( x) = sec 2 x
⎛ π⎞
f ′⎜ − ⎟ = 2
⎝ 4⎠
f ′′( x) = 2 sec 2 x tan x
⎛ π⎞
f ′′⎜ − ⎟ = − 4
⎝ 4⎠
f ′′′( x) = 4 sec 2 x tan 2 x + 2 sec4 x
⎛ π⎞
f ′′′⎜ − ⎟ = 16
⎝ 4⎠
2
π⎞
π⎞
8⎛
π⎞
⎛
⎛
P3 ( x ) = −1 + 2⎜ x + ⎟ − 2⎜ x + ⎟ + ⎜ x + ⎟
4
4
3
4⎠
⎝
⎠
⎝
⎠
⎝
67. Because
(95π )9
1809 ⋅ 9!
3
< 0.001, use four terms.
(95π ) + (95π ) − (95π ) ≈ 0.99594
95π
⎛ 95π ⎞
−
sin 95° = sin ⎜
⎟ ≈
180
18033!
18055!
18077!
⎝ 180 ⎠
3
68. Because
(0.75)6
6!
cos(0.75) ≈ 1 −
(0.75)
5
7
< 0.001, use three terms.
(0.75)
2
+
2!
(0.75)
4
≈ 0.7319
4!
15
69. Because
< 0.001, use 14 terms.
15
ln (1.75) ≈ (0.75) −
70. Because
(0.25)
5!
(0.75)2
2
+
(0.75)3
3
−
(0.75)4
4
+
(0.75)5
5
−
(0.75)6
6
+
−
(0.75)14
14
≈ 0.559062
5
< 0.001, use five terms.
e −0.25 ≈ 1 − 0.25 +
(0.25)2
2!
−
(0.25)3
3!
+
(0.25)4
4!
≈ 0.779
© 2010 Brooks/Cole, Cengage Learning
384
Chapter 9
Infinite Series
71. f ( x) = cos x,
Rn ( x) =
c = 0
f (n +1) ( z )
(n
+ 1)!
x n +1
f (n + 1) ( z ) ≤ 1 ⇒ Rn ( x) ≤
(a) Rn ( x) ≤
(0.5)
(n + 1)!
(n
x
n
( + 1)!
n→∞
(0.5)
(n + 1)!
Because the series converges when x = 1 and when
x = 3, the interval of convergence is [1, 3].
< 0.001
∞
76.
∑
3n ( x − 2)
< 0.0001
lim
n→∞
R =
This inequality is true for n = 10.
x2
x4
+
2!
4!
P6 ( x) = 1 −
x2
x4
x6
+
−
2!
4!
6!
x2
x4
x6
x8
x10
P10 ( x) = 1 −
+
−
+
−
2!
4!
6!
8!
10!
∞
77.
n→∞
P10
−10
Geometric series which converges only if x 10 < 1 or
−10 < x < 10.
∞
∑ ( 2 x)
n
5
7
and diverges at , the
3
3
n
(n + 1)!( x − 2)
un + 1
= lim
n
n→∞
un
n!( x − 2)
n +1
= ∞
which implies that the series converges only at the center
x = 2.
10
f
n=0
3 ( x − 2)
n=0
P4
74.
∑ n!( x − 2)
lim
n
n
n
⎡5 7 ⎞
interval of convergence is ⎢ , ⎟.
⎣3 3 ⎠
10
⎛x⎞
⋅
1
3
Because the series converges at
P4 ( x) = 1 −
∞
n +1
Center: 2
72. f ( x) = cos x
∑ ⎜⎝ 10 ⎟⎠
3n + 1 ( x − 2)
un + 1
= lim
n→∞
un
n +1
= 3 x − 2
2n +1
(d) Rn ( x) ≤
< 0.0001
(n + 1)!
73.
n
n
n =1
This inequality is true for n = 5.
P6
(n + 1)2
n
n
(−1) ( x − 2)
Center: 2
n +1
−10
⋅
R =1
< 0.001
This inequality is true for n = 6.
(c) Rn ( x) ≤
n +1
= x − 2
n +1
+ 1)!
(−1) ( x − 2)
un + 1
= lim
2
n
→
∞
un
( n + 2)
n +1
n +1
(1)
∑
lim
n +1
This inequality is true for n = 4.
(b) Rn ( x) ≤
(−1)n ( x − 2)n
n=0
(n + 1)2
∞
75.
∞
78.
∑
n=0
(x
− 2)
n
2n
=
∞
⎛ x − 2⎞
⎟
2 ⎠
n=0
∑ ⎜⎝
n
Geometric series which converges only if
x − 2
<1
2
or
0 < x < 4.
n
n=0
Geometric series which converges only if 2 x < 1 or
− 12 < x < 12 .
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
y =
79.
∞
∑ (−1)
x 2n
n
4 ( n!)
n
n=0
(−1) (2n) x 2 n −1
∑ 4n n! 2
n =1
( )
y′′ =
∑
∞
x 2 y′′ + xy′ + x 2 y =
∞
∑
(−1)n +1 (2n
⎡
n=0
⎣
∞
∞
⎡ ( −1)n + 1 4( n + 1)2
n +1
⎣ 4 ⎡⎣( n + 1)!⎤⎦
⎡ ( −1)n + 11
∑ ⎢⎢ 4n
∞
∑
∞
∑
⎣
∞
∑
(n!)
∞
∑
∞
∑
∞
∑
n
∑
∑
=
∑
n =1
81.
n=0
x 2n + 2
4n ( n!)
2
(−1)n ⎤⎥x 2 n + 2
2
4n ( n!) ⎥
⎦
⎤
⎥x 2 n + 2
4n ( n!) ⎥
⎦
1
2
⎤
⎥x 2 n + 2 = 0
4n ( n!) ⎦⎥
1
n
2
(−3) (2n + 2) x 2n +1
2n + 1 ( n + 1)!
n=0
n +1
∞
∑
+ 2)( 2n + 1) x 2 n
(−3) (2n
+ 2)( 2n + 1) x 2 n
2n + 1 ( n + 1)!
n +1
2
(−1)
(n
+ 1)!
( 2n
+ 2) x
n +1
n +1 n +1
3
2 n!
(−1)
n n +1 2n
3
x
2n n!
(−1) 3n +1 x 2 n
n
n
2 n!
(−1)
x
2 n!
n n +1 2 n
3
∑
(−1)
x
(−1)
+
( 2n )
+
∞
∑
∞
∑
n =1
[−2n
3
n +1 n + 2
3
(−1)
x
2n + 2
+
2 n!
n=0
(−2n)
( 2 n + 2) x 2 n + 2
2n + 1 ( n + 1)!
n +1 n + 2
n
∞
∑
(−1)
3
∞
∑
n=0
n +1 n + 2
3
x
(−1)
+
∞
∑
n=0
(−1)
n n +1 2 n
3
x
n
2 n!
n n +1 2n
3
x
n
2 n!
2n + 2
2n n!
n=0
n +1 n + 2
x 2n + 2
n
2 n!
n=0
3
2n n!
+
∞
⎡−
⎣ ( 2n + 1) + 1⎤⎦ +
n +1 n +1 2 n
n
(−1)
∞
∑
+
n=0
2n
n
n =1
∞
n
(−3)n +1 (2n
n=0
∞
=
2n n!
n=0
∞
+ ( −1)
2
+ ( −1)
2
(−3) (2n) x 2n −1
n=0
=
n
2n n!
n=0
=
∞
∑ (−1)
(−3)n x 2 n
n=0
=
2
+
∑ ⎢⎢
n =1
=
4n + 1 ⎡⎣( n + 1)!⎤⎦
(−1)n +1(2n + 2)
2
4n + 1 ⎡⎣( n + 1)!⎤⎦
+
=
n=0
y′′ + 3 xy′ + 3 y =
+ 2)( 2n + 1)
+
∞
n=0
y′′ =
( 2n
(−1)n +1 ( 2n + 2) x 2n + 2
2
n=0
4n + 1 ⎡⎣( n + 1)!⎤⎦
∞
∑
+
2
⎡ ( −1)n + 1 ( 2n + 2)( 2n + 1 + 1)
1 ⎤⎥ 2 n + 2
n
+ ( −1)
x
∑ ⎢⎢
2
2
n
4 ( n!) ⎥⎦
n=0
4n + 1 ⎡⎣( n + 1)!⎤⎦
⎣
=
y′ =
+ 2)( 2n + 1) x 2 n + 2
=
n=0
y =
n +1
2
2
4n + 1 ⎡⎣( n + 1)!⎤⎦
∞
+ 1)!⎤⎦
+ 2)( 2n + 1) x 2 n
( −1)n +1 (2n
∑ ⎢⎢(−1)
+ 2) x 2 n + 1
n +1
∞
=
4n + 1 ⎡⎣( n + 1)!⎤⎦
n=0
=
(−1) (2n
∑ n +1
n=0
4 ⎣⎡( n
n
n=0
80.
2
y′ =
∞
385
(−1)
2
n n +1 2 n
n −1
3
x
(n
− 1)!
⋅
2n
2n
+ 2n] = 0
2
23
a
=
=
3− x
1 − ( x 3)
1− r
∞
2⎛ x ⎞
∑ 3 ⎜⎝ 3 ⎟⎠
n=0
n
=
∞
2xn
∑ 3n +1
n=0
© 2010 Brooks/Cole, Cengage Learning
386
82.
Chapter 9
Infinite Series
3
32
32
a
=
=
=
2+ x
1 + ( x 2)
1 − (− x 2)
1− r
∞
3⎛ x ⎞
∑ 2 ⎜⎝ − 2 ⎟⎠
n
∞
∑
=
n=0
n=0
(−1)
n
3x n
2n + 1
2
. Power series
3− x
83. g ( x) =
∞
2 ⎛ x⎞
∑ 3 n⎜⎝ 3 ⎟⎠
Derivative:
n −1
n =1
∞
2⎛ x ⎞
∑ 3 ⎜⎝ 3 ⎟⎠
n
n=0
⎛1⎞
⎜ ⎟ =
⎝ 3⎠
∞
n −1
2 ⎛ x⎞
∑ 9 n⎜⎝ 3 ⎟⎠
∞
⎛ x⎞
2
∑ 9 (n + 1)⎜⎝ 3 ⎟⎠
=
n =1
n
n=0
( −1)n 3x n +1
n +1
n = 0 ( n + 1) 2
∞
84. Integral: ∑
85. 1 +
2
4
8 3
x + x2 +
x +
3
9
27
86. 8 − 2( x − 3) +
87.
=
∞
⎛ 2x ⎞
∑ ⎜⎝ 3 ⎟⎠
n
=
n=0
⎛ 3 3⎞
⎜− , ⎟
⎝ 2 2⎠
1
3
,
=
1 − ( 2 x 3)
3 − 2x
⎡−( x − 3) ⎤
8
⎥ =
4
1 − ⎡−
n=0 ⎣
⎦
⎣ ( x − 3) 4⎤⎦
32
32
=
=
, ( −1, 7)
4 + ( x − 3)
1+ x
1
1
( x − 3)2 − ( x − 3)3 +
2
8
n
∞
∑ 8⎢
=
f ( x) = sin x
f ′( x) = cos x
f ′′( x) = −sin x
f ′′′( x) = −cos x,
sin ( x) =
∞
∑
f (n) ( x) ⎣⎡ x − (3π 4)⎤⎦
n!
n=0
88.
2 ∞ ( −1)
∑
2 n=0
2
2⎛
3π ⎞
2 ⎛
3π ⎞
⎜x −
⎟ −
⎜x −
⎟ +
2 ⎝
4 ⎠ 2 ⋅ 2!⎝
4 ⎠
2
−
2
=
n
=
n( n + 1) 2
⎡⎣ x − (3π 4)⎤⎦
n!
n
f ( x) = cos x
f ′( x) = −sin x
f ′′( x) = −cos x
f ′′′( x) = sin x
cos x =
∞
∑
n=0
π⎞
2 ⎢⎡
⎛
1 + ⎜x + ⎟ +
2 ⎢
4⎠
⎝
⎣
=
( )
89. 3x = eln(3)
3x =
∞
∑
n=0
f (n) ( −π 4) ⎡⎣ x + (π 4)⎤⎦
n!
x
∞
∑
n
2
+
2
=
(−1)⎣⎡ (
n n + 1)⎦⎤ 2
⎡⎣ x + (π 4)⎤⎦
(n + 1)!
n =1
= e x ln(3) and because e x =
∞
∑
n=0
( x ln 3)
n!
n
= 1 + x ln 3 +
2
x [ln 3]
2!
4
n +1
⎤
⎥
⎥
⎦
xn
, you have
n!
2
2
3
π⎞
π⎞
π⎞
π⎞
2⎛
2 ⎛
2 ⎛
2 ⎛
⎜x + ⎟ −
⎜x + ⎟ −
⎜x + ⎟ +
⎜x + ⎟ +
2 ⎝
4 ⎠ 2 ⋅ 2!⎝
4⎠
2 ⋅ 3!⎝
4⎠
2 ⋅ 4!⎝
4⎠
x3 [ln 3]
3
+
3!
x 4 [ln 3]
4
+
4!
+
.
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
90.
387
f ( x) = csc( x)
f ′( x) = −csc( x) cot ( x)
f ′′( x) = csc3 ( x) + csc( x) cot 2 ( x)
f ′′′( x) = −5 csc3 ( x) cot ( x) − csc( x) cot 3 ( x)
f (4) ( x) = 5 csc5 ( x) + 15 csc3 ( x) cot 2 ( x) + csc( x) cot 4 ( x)
csc( x) =
f (n) (π 2) ⎡⎣ x − (π 2)⎤⎦
n!
∞
∑
n=0
91.
f ( x) =
n
2
= 1+
4
π⎞
π⎞
1⎛
5⎛
⎜x − ⎟ + ⎜x − ⎟ +
2!⎝
2⎠
4!⎝
2⎠
1
x
1
x2
2
f ′′( x) = 3
x
6
f ′′′( x) = − 4 ,
x
f ′( x) = −
1
=
x
92.
∞
∑
n=0
f (n) ( −1)( x + 1)
n!
n
=
− n!( x + 1)
∑ n!
n=0
∞
n
∞
= − ∑ ( x + 1) , − 2 < x < 0
n
n=0
f ( x) = x1 2
1 −1 2
x
2
⎛ 1 ⎞⎛ 1 ⎞
f ′′( x) = −⎜ ⎟⎜ ⎟ x −3 2
⎝ 2 ⎠⎝ 2 ⎠
f ′( x) =
⎛ 1 ⎞⎛ 1 ⎞⎛ 3 ⎞
f ′′′( x) = ⎜ ⎟⎜ ⎟⎜ ⎟ x −5 2
⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠
⎛ 1 ⎞⎛ 1 ⎞⎛ 3 ⎞⎛ 5 ⎞
f (4) ( x) = −⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟ x −7 2 ,
⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠⎝ 2 ⎠
x =
∞
∑
n=0
f (n) ( 4)( x − 4)
n!
= 2+
93.
(1 + x)k
(1 + x)
15
− 4)
+
22
2
3
1 ⋅ 3 ⋅ 5 ( 2n − 3)( x − 4)
23n −1 n!
n +1
4
n
= 1+
x
+
5
∞
∑
(−1)n +1 4 ⋅ 9 ⋅ 14 (5n − 6) x n
5n n!
n=2
= 1+
x
2 2
6 3
−
x +
x −
5 25
125
−3
h′′( x) = 12(1 + x)
−4
−5
h′′′( x) = −60(1 + x)
h(4) ( x) = 360(1 + x)
−6
−7
h(5) ( x) = −2520(1 + x)
(1 + x)
1 ⋅ 3( x − 4)
1 ⋅ 3 ⋅ 5( x − 4)
− 4) ( x − 4)
−
+
−
+
22
252!
283!
2114!
1
1 ⋅ 4 x 2 1 ⋅ 4 ⋅ 9 x3
x− 2
+
−
5
5 2!
533!
h′( x) = −3(1 + x)
3
n=2
(−1)
(x
k ( k − 1) x 2 k ( k − 1)( k − 2) x3
+
+
2!
3!
x (1 5)( −4 5) x 2 1 5( −4 5)( −9 5) x3
= 1+ +
+
+
5
2!
3!
h( x) = (1 + x)
1
∞
∑
= 2+
= 1 + kx +
= 1+
94.
(x
n
= 1 − 3x +
−8
12 x 2
60 x3
360 x 4
2520 x5
−
+
−
+
2!
3!
4!
5!
=
∞
(−1) (n + 2)!x n
n=0
2n!
∑
n
=
∞
(−1) (n + 2)( n + 1) x n
n=0
2
∑
n
© 2010 Brooks/Cole, Cengage Learning
388
95.
Chapter 9
Infinite Series
∞
∑ (−1)
ln x =
( x − 1)
n +1
96.
n +1⎛
∞
∑ (−1)
⎜
⎝
n =1
ln x =
∞
∑ (−1)
(5 4) − 1⎞
n +1
∞
∑ (−1)
=
∑ (−1)
n +1 ⎛
⎜
⎝
n =1
∞
n +1
n =1
97.
ex =
∞
∑
n=0
e1 2 =
∞
(1 2)
n=0
n!
∑
1
102.
xn
,
n!
1+ x
3
n
=
n
⎟
⎠
n
∞
1
∑ 2n n!
−1 2
f ′′( x) = 4e
∞
2n
∑ 3n n!
=
≈ 1.9477
n=0
∞
x 2n
∑ (−1) (2n)!,
n=0
n
∞
∑ (−1)
n
n=0
∞
−∞ < x < ∞
22 n
≈ 0.7859
3 ( 2n)!
2n
x 2 n +1
∑ (−1) (2n + 1)!,
n=0
n
∞
∑ (−1)
n
n=0
2n +1
3
−∞ < x < ∞
1
≈ 0.3272
(2n + 1)!
101. The series in Exercise 45 converges very slowly because
the terms approach 0 at a slow rate.
1
2
,k = −
=1−
x3
+
2
∞
∑
n=2
(−1) 1 ⋅ 3 (2n − 1) x3n
n
2n n!
f (0) = 1
f ′(0) = 2
2x
2x
f ′′(0) = 4
f ′′′(0) = 8
P( x) = 1 + 2 x +
∞
n!
sin x =
−1 2 −3 2
−1 2 − 3 2 − 5 2
1 3
( x ) + ( )(2! ) x6 + ( )( 3! )( ) x9 +
2
f ( x) = e 2 x
n=0
⎛1⎞
sin ⎜ ⎟ =
⎝ 3⎠
≈ 1.6487
−∞ < x < ∞
(2 3) n
cos x =
100.
xn
,
n!
∞
∑
⎛ 2⎞
cos⎜ ⎟ =
⎝ 3⎠
n
1
≈ 0.1823
5n n
f ′′′( x) = 8e 2 x
4x2
8 x3
4
+
= 1 + 2 x + 2 x 2 + x3
2!
3!
3
xn
∑ n! , e2 x
∞
( 2 x)n
n=0
n!
=
n=0
∑
P( x) = 1 + 2 x + 2 x 2 +
4 3
x
3
⎛
x2
+
(c) e x ⋅ e x = ⎜1 + x +
2!
⎝
⎞⎛
x2
+
⎟⎜1 + x +
2!
⎠⎝
4
P( x) = 1 + 2 x + 2 x 2 + x3
3
104. (a)
e2 3 =
99.
0 < x ≤ 2
(6 5) − 1 ⎞
= (1 + x3 )
ex =
n =1
1
≈ 0.2231
4n n
n=0
f ′( x) = 2e
(b)
n +1
∞
∑
n=0
−∞ < x < ∞
= 1−
103. (a)
∑ (−1)
− 1)
,
n
n =1
⎛6⎞
ln ⎜ ⎟ =
⎝5⎠
∞
⎟ =
⎠
n
(x
n
ex =
98.
0< x≤2
,
n
n =1
⎛ 5⎞
ln⎜ ⎟ =
⎝ 4⎠
n
f ( x) = sin 2 x
f ( 0) = 0
f ′( x) = 2 cos 2 x
f ′(0) = 2
f ′′( x) = −4 sin 2 x
f ′′(0) = 0
f ′′′( x) = −8 cos2 x
f ′′′(0) = −8
f (4) ( x) = 16 sin 2 x
f ( 4 ) ( 0) = 0
f (5) ( x) = 32 cos 2 x
f (5) (0) = 32
f (6) ( x) = −64 sin 2 x
f ( 6 ) ( 0) = 0
f (7) ( x) = −128 cos 2 x
f (7) (0) = −128
sin 2 x = 0 + 2 x +
⎞
⎟
⎠
0x2
0x4
32 x5
0 x 6 128 x 7
8 x3
−
+
+
+
−
+
2!
3!
4!
5!
6!
7!
= 2x −
8 7
4 3
4 5
x −
x +
x +
315
3
15
© 2010 Brooks/Cole, Cengage Learning
Review Exercises for Chapter 9
(b)
∞
(−1)n x 2n +1
∞
(−1)n (2 x)2n +1
sin x =
∑
n = 0 ( 2n + 1)!
sin 2 x =
∑ (2n + 1)!
n=0
( 2 x )3
= 2x −
+
3!
8 x3
32 x5 128 x 7
+
−
+
6
120
5040
= 2x −
( 2 x )5
5!
= 2x −
−
( 2 x )7
7!
389
+
4 3
4 5
8 7
x +
x −
x +
3
15
315
(c) sin 2 x = 2 sin x cos x
⎛
x3
x5
x7
= 2⎜ x −
+
−
+
6
120 5040
⎝
⎞⎛
x2
x4
x6
+
−
+
⎟⎜1 −
2
24 720
⎠⎝
⎞
⎟
⎠
⎡
⎛ x3
x3 ⎞ ⎛ x5
x5
x5 ⎞ ⎛ x7
x7
x7
x7 ⎞
= 2⎢ x + ⎜ −
−
+
+
−
−
−
⎟+⎜
⎟ + ⎜−
⎟+
6 ⎠ ⎝ 24 12 120 ⎠ ⎝ 720 144
240 5040 ⎠
⎝ 2
⎣⎢
⎛
2 x3
2 x5
4 x7
= 2⎜ x −
+
−
+
3
15
315
⎝
105.
sin t =
(−1)n t 2 n +1
∑ (2n + 1)!
n=0
sin t
=
t
(−1)n t 2n
∑ (2n + 1)!
n=0
⎞
4 3
4 5
8 7
x −
x +
⎟ = 2x − x +
3
15
315
⎠
∞
108. et =
∫0
et − 1 =
∞
e −1
=
t
x
∫0
(−1)n x 2n +1
∞
tn
∞
tn
∑ n!
n =1
t
⎡∞
(−1)n t 2n +1 ⎤⎥
sin t
dt = ⎢ ∑
t
⎢⎣n = 0 ( 2n + 1)( 2n + 1)!⎥⎦ 0
=
∞
∑ n!
n=0
x
x
∑ (2n + 1)(2n + 1)!
t n −1
n = 1 n!
∞
∑
x
⎡ ∞ tn ⎤
et − 1
dt = ⎢∑
⎥ =
t
⎣n =1 n ⋅ n!⎦ 0
n=0
cos t =
106.
t
=
2
cos
109.
(−1)n t 2 n
∞
∑
n = 0 ( 2n)!
∑ 22n (2n)!
n=0
x
x
∫0
⎡∞
(−1)n t n +1 ⎤⎥
t
dt = ⎢ ∑ 2 n
2
⎢⎣n = 0 2 ( 2n)!( n + 1) ⎥⎦ 0
cos
=
∑ 22n (2n)!(n + 1)
1
=
1+t
ln (1 + t ) =
ln (t + 1)
t
x
∫0
ln (t + 1)
t
=
∞
∑ (−1)
n n
t
110.
n=0
1
∫ 1 + t dt =
∞
∑
n=0
(−1)
n =1
x 3 x5 x 7 x9
+
−
+
−
3
5
7
9
⎛ 1 ⎞
⎜
⎟
arctan x
2 x
1 + x2 ⎠
= lim ⎝
= lim
= 0.
lim
2
+
+
+
1
⎞
x→0
x→0 ⎛
x→0 1 + x
x
⎜
⎟
⎝2 x ⎠
n=0
107.
xn
By L'Hôpital's Rule,
(−1)n x n +1
∞
arctan x = x −
∞
∑ n ⋅ n!
arctan x
x5 2 x9 2 x13 2 x17 2
= x−
+
−
+
−
3
5
7
9
x
arctan x
=0
lim
x → 0+
x
(−1)n t n
∞
⎤
⎥
⎦⎥
∞
∑
(−1)
n=0
t
n +1
t
n +1
x
∞
∑
n=0
(−1) x n +1
n
(n
x3
1 ⋅ 3x5
1 ⋅ 3 ⋅ 5 x7
+
+
+
2⋅3 2⋅4⋅5 2⋅4⋅6⋅7
arcsin x
x2
1 ⋅ 3x 4
1 ⋅ 3 ⋅ 5 x6
=1+
+
+
+
x
2⋅3 2⋅4⋅5 2⋅4⋅6⋅7
arcsin x
lim
=1
x→0
x
n n +1
n n
⎡ ∞ ( −1)n t n + 1 ⎤
⎥ =
dt = ⎢ ∑
2
⎢⎣n = 0 ( n + 1) ⎥⎦
0
arcsin x = x +
+ 1)
2
By L'Hôpital's Rule,
⎛
⎜
arcsin x
lim
= lim ⎝
x→0
x→0
x
⎞
⎟
1− x ⎠
= 1.
1
1
2
© 2010 Brooks/Cole, Cengage Learning
390
Chapter 9
Infinite Series
Problem Solving for Chapter 9
⎛1⎞
⎛1⎞
⎛ 1⎞
1. (a) 1⎜ ⎟ + 2⎜ ⎟ + 4⎜ ⎟ +
3
9
⎝ ⎠
⎝ ⎠
⎝ 27 ⎠
=
∞
1⎛ 2 ⎞
∑ 3⎜⎝ 3 ⎟⎠
n
=
n=0
13
=1
1 − ( 2 3)
1 2
(b) 0, , , 1, etc.
3 3
(c) lim Cn = 1 −
n →∞
∞
1⎛ 2 ⎞
∑ 3⎜⎝ 3 ⎟⎠
n
= 1−1 = 0
n=0
2. (a) Let ε > 0 be given. lim a2 n = L means there exists M 1 such that a2n − L < ε for n > M 1. lim a2 n + 1 = L means
n→∞
n→∞
there exists M 2 such that a2 n + 1 − L < ε for n > M 2 . Let M = max{2 M 1 , 2 M 2 + 1}. Then for n > M , and n = 2m
even, you have 2m > M > 2 M 1 ⇒ m > M 1 ⇒ a2 m − L < ε . And for n > M , n = 2m + 1 odd, you have
2m + 1 > M > 2 M 2 + 1 ⇒ m > M 2 ⇒ a2 n + 1 − L < ε . So, lim an = L.
n→∞
1
.
(b) a1 = 1, an + 1 = 1 +
1 + an
1
1
3
a2 = 1 +
=1+
=
= 1.5
1 + a1
1+1
2
1
1
7
a3 = 1 +
=1+
=
= 1.4
1 + a2
1 + ( 3 2)
5
a4 =
a5 =
a6 =
a7 =
a8 =
17
= 1.416
12
41
≈ 1.4140
29
99
≈ 1.41429
70
239
≈ 1.414201
169
577
≈ 1.414216
408
Using mathematical induction, you can show that the odd terms are increasing and the even terms are decreasing. Both
sequences are bounded in [1, 2]. So, both sequences converge.
Let lim a2 n = L. Then lim a2 n + 2 = L, and
n→∞
n→∞
an + 2 = 1 +
1
=1+
1 + an +1
⇒ a2 n + 2 =
4 + 3a2 n
3 + 2a2 n
So, L =
3. Let S =
4 + 3L
⇒ 2 L2 = 4 ⇒ L =
3 + 2L
∞
∑
1
n =1
1
=1+
⎡
1 ⎤
1 + ⎢1 +
1+
1 + an ⎥⎦
⎣
( 2n
− 1)
2
=
1
1
1
+ 2 + 2 +
12
3
5
1
1
1 + an
4 + 3an
= 1+
= 1+
=
3 + 2an
3 + 2an
⎡ 2 + an ⎤
⎛ 3 + 2an ⎞
⎜
⎟
⎢1 + a ⎥
n ⎦
⎣
⎝ 1 + an ⎠
2. Similarly, lim a2 n +1 =
n→∞
2. So by part (a), lim an = L =
n→∞
2
.
Then
π2
6
1
1
1
1
+ 2 + 2 + 2 +
12
2
3
4
1
1
= S + 2 + 2 +
2
4
=
= S +
So, S =
π2
6
1⎡
1
1
1+ 2 + 2 +
22 ⎢⎣
2
3
−
1 ⎛π 2 ⎞
⎤
⎥ = S + 22 ⎜ 6 ⎟.
⎦
⎝ ⎠
1π2
π 2 ⎛ 3⎞ π 2
=
.
⎜ ⎟ =
4 6
6 ⎝ 4⎠
8
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 9
4. If there are n rows, then an =
n( n + 1)
2
5. (a) Position the three blocks as indicated in the figure.
The bottom block extends 1 6 over the edge of the
.
table, the middle block extends 1 4 over the edge of
1⎛ 3 ⎞
3
1
For one circle, a1 = 1 and r1 = ⎜⎜
=
.
⎟=
3 ⎝ 2 ⎟⎠
6
2 3
the bottom block, and the top block extends 1 2
over the edge of the middle block.
The centers of gravity are located at
1
1
1
bottom block: −
= −
6
2
3
1
1
1
1
= −
middle block: + −
6
4
2
12
1
1
1
1
5
top block: + + −
=
.
6
4
2
2
12
The center of gravity of the top 2 blocks is
1
5⎞
⎛ 1
+
⎜−
⎟ 2 = , which lies over the bottom
12
12
6
⎝
⎠
block. The center of gravity of the 3 blocks is
1
5⎞
⎛ 1
+
⎜− −
⎟ 3 = 0 which lies over the table.
⎝ 3 12 12 ⎠
So, the far edge of the top block lies
1
1
1
11
beyond the edge of the table.
+ +
=
6
4
2
12
3
2
1
r1
1
2
For three circles, a2 = 3 and 1 = 2 3r2 + 2r2
r2 =
1
.
2+ 2 3
2r2
r2
3 r2
1
For six circles, a3 = 6 and 1 = 2 3r3 + 4r3
r3 =
391
1
.
2 3 + 4
1
2
1
4
1
6
0 1 5
6 12
2r3
1
(b) Yes. If there are n blocks, then the edge of the top block
n
1
lies ∑ from the edge of the table. Using 4 blocks,
2
i =1 i
r3
3 r3
Continuing this pattern, rn =
1
.
2 3 + 2( n − 1)
4
2
∑ an x n
1
∑ 2i
i =1
⎛
⎞ n(n + 1)
1
Total Area = (π rn2 )an = π ⎜
⎟
⎜ 2 3 + 2(n − 1) ⎟
2
⎝
⎠
n(n + 1)
π
An =
2 ⎡2 3 + 2(n − 1)⎤ 2
⎣
⎦
π 1 π
lim An = ⋅ =
n →∞
2 4
8
6. (a)
11
12
=
1
1
1 1
25
+ + +
=
2
4
6 8
24
which shows that the top block extends beyond the
table.
(c) The blocks can extend any distance beyond the table
because the series diverges:
∞
1
1 ∞ 1
∑ 2i = 2 ∑ i = ∞.
i =1
i =1
= 1 + 2 x + 3x 2 + x3 + 2 x 4 + 3x5 +
= (1 + x3 + x 6 +
= (1 + x3 + x 6 +
) + 2( x + x 4 + x7 + ) + 3( x 2 + x5 + x8 + )
)(1 + 2 x + 3x 2 ) = (1 + 2 x + 3x 2 )1 −1 x3
R = 1 because each series in the second line has R = 1.
(b)
∑ an x n
= ( a0 + a1 x +
= a0 (1 + x +
p
= ( a0 + a1 x +
+ a p −1 x p −1 ) + ( a0 x p + a1 x p +1 +
) + a1x(1 + x + ) +
+ a p −1 x p −1 )(1 + x p +
p
)+
+ a p −1 x (1 + x p + )
) = (a0 + a1x + + a p −1x p −1 )1 −1 x p
p −1
R =1
(Assume all an > 0.)
© 2010 Brooks/Cole, Cengage Learning
392
Chapter 9
Infinite Series
b
a
b
+
− +
2
3
4
7. a −
=
(−1) (2a)
n =1
2n
∞
If a ≠ b, ∑
n +1
+ b) + ( a − b )
2n
∞
= a∑
(−1)
+ b)
n +1
2n
+
n +1
n
n =1
(−1) (a
n =1
(−1)n +1(a
n =1
∞
If a = b, ∑
∞
∑
converges conditionally.
∞
a −b
diverges.
n = 1 2n
∑
No values of a and b give absolute convergence. a = b implies conditional convergence.
x2
+
2!
ex = 1 + x +
8. (a)
xe =
x
∞
∑
n=0
=
∞
12. (a) a1 = 3.0
xn
∑ n!
a2 ≈ 1.73205
n=0
x n +1
n!
a3 ≈ 2.17533
xn + 2
=
−
+
=
xe
dx
xe
e
C
∑
∫
n = 0 ( n + 2) n!
x
x
∞
x
Letting x = 0, you have C = 1. Letting x = 1,
∞
1
1
e −e +1= ∑
=
+
2
n = 0 ( n + 2) n!
∞
1
∑ (n + 2)n!.
n =1
∞
1
1
So, ∑
= .
2
n = 1 ( n + 2) n!
(b) Differentiating, xe x + e x =
∞
∑
(n
n=0
+ 1) x n
n!
.
n +1
≈ 5.4366.
Letting x = 1, 2e = ∑
n!
n=0
2
x
+
2!
x4
= 1 + x2 +
+
2!
ex = 1 + x +
ex
2
+
x12
+
6!
f (12) (0)
1
12!
=
⇒ f (12) (0) =
= 665,280
12!
6!
6!
3π
∫ 2π
sin x
dx = a1 − a2 + a3 − a4 + .
x
Because lim an = 0 and an + 1 < an , this series converges.
∞
∫0
n→∞
∞
2
1
∞
dx = [ln ln x] 2 diverges.
x ln x
If
1
∞
2
x (ln x)
p
⎡ (ln b)1− p (ln 2)1− p ⎤
−
⎥
dx = lim ⎢
b →∞ ⎢ 1 − p
1− p ⎥
⎣
⎦
converges.
If p < 1, diverges.
(b)
∞
1
∑ n ln n 2
n=4
lim an =
( )
=
1 ∞
1
diverges by part (a).
∑
2 n = 4 n ln n
1+
13
2
n→∞
[See part (b) for proof.]
(b) Use mathematical induction to show the sequence is
increasing. Clearly,
a + a1 =
a a >
a = a1. Now
assume an > an −1. Then
an + a > an −1 + a
an + a >
an −1 + a
an + 1 > an .
Use mathematical induction to show that the
sequence is bounded above by a. Clearly,
a1 = a < a. Now assume an < a. Then
a > an and a − 1 > 1 implies
a( a − 1) > an (1)
a 2 − a > an
a 2 > an + a
a >
an + a = an + 1.
So, the sequence converges to some number L. To
find L, assume an + 1 ≈ an ≈ L :
L =
L =
11. (a) If p = 1, ∫
p > 1, ∫
a6 ≈ 2.30146
2π
∫0
10. Let a1 =
a3 =
sin x
sin x
dx, a2 = − ∫
dx,
π
x
x
sin x
dx, etc. Then,
x
π
a5 ≈ 2.29672
a2 =
∞
9.
a4 ≈ 2.27493
a + L ⇒ L2 = a + L ⇒ L2 − L − a = 0
1±
So, L =
1 + 4a
.
2
1+
1 + 4a
.
2
13. Let bn = an r n .
(bn )1 n
Lr <
= ( an r n )
1n
= a1n n ⋅ r → Lr as n → ∞.
1
r = 1.
r
By the Root Test,
∑ bn converges ⇒ ∑ an r n
converges.
© 2010 Brooks/Cole, Cengage Learning
Problem Solving for Chapter 9
14. (a)
∞
∑
n =1
1
2
n + (−1)n
=
1
1
1
1
1
+
+
+
+
+
21−1 22 +1 23−1 24 +1 25 −1
1
1
⎡
17. (a) Height = 2 ⎢1 +
+
+
2
3
⎣
∞
1
=1
20
1 9
= 1+ =
8 8
9 1 11
= + =
8 4
8
11 1
45
=
+
=
8 32 32
45 1
47
=
+
=
32 16 32
= 2∑
S1 =
S1
S3
S4
S5
n =1
(c)
2( −1)
2n + (−1)
an + 1
=
=
n + 1) + ( −1)n + 1
1 + (−1)n + 1
(
an
2
2
1
1
This sequence is , 2, , 2, … which diverges.
8
8
n
1n
1
2n + (−1)
n
⎛
⎞
1
=⎜
⎜ n (−1)n ⎟⎟
⎝2 ⋅ 2
⎠
{ }
converges because 2
and
n
1 2 → 1 and
(−1)n
n
1
=
2⋅
n
2(−1)
n
1
1
= , 2, , 2, …
2
2
2 → 1.
→
4 ⎡
1
1
π 1 + 32 + 32 +
3 ⎢⎣
2
3
=
4 ∞ 1
converges.
π∑
3 n =1 n 3 2
∞
18. (a)
1
∑ 2n
1
1
=
=
0.99
1 − 0.01
∞
∑ (0.01)
n
n=0
= 1 + 0.01 + (0.01) +
2
1
1
=
=
0.98
1 − 0.02
∞
∑ (0.02)
⎤
⎥
⎦
1 ∞ 1
∑ diverges (harmonic)
2 n =1 n
(b) Let f ( x) = sin x. By the Mean Value Theorem,
1
<1
2
f ( x) − f ( y ) = f ′(c) x − y
= cos(c) x − y ≤ x − y ,
where c is between x and y. So,
⎛ 1 ⎞
⎛ 1 ⎞
0 ≤ sin ⎜ ⎟ − sin ⎜
⎟
⎝ 2n ⎠
⎝ 2n + 1 ⎠
1
1
1
−
=
2n 2n + 1
2n( 2n + 1)
Because
∞
1
∑ 2n(2n + 1) converges, the Comparison
n =1
Theorem tells us that
∞
⎡ ⎛ 1 ⎞
⎛ 1 ⎞⎤
∑ ⎢sin⎜⎝ 2n ⎟⎠ − sin⎜⎝ 2n + 1 ⎟⎠⎥ converges.
⎦
n =1 ⎣
= 1.010101
(b)
=
n =1
≤
15. (a)
∞
1
⎤
4
=
π
= ∞
∑
⎥
⎦
n =1 n
(c) W =
n
⎤
⎥
⎦
1
1
⎛
⎞
= ∞ ⎜ p -series, p =
< 1⎟
n1 2
2
⎝
⎠
1 1
⎡
(b) S = 4π ⎢1 + + +
2 3
⎣
n
(b)
393
n
n=0
= 1 + 0.02 + (0.02) +
2
= 1 + 0.02 + 0.0004 +
= 1.0204081632
16. S6 = 130 + 70 + 40 = 240
S7 = 240 + 130 + 70 = 440
S8 = 440 + 240 + 130 = 810
S9 = 810 + 440 + 240 = 1490
S10 = 1490 + 810 + 440 = 2740
© 2010 Brooks/Cole, Cengage Learning