C H A P T E R
Infinite Series
9
Section 9.1
Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . 233
Section 9.2
Series and Convergence . . . . . . . . . . . . . . . . . . . 247
Section 9.3
The Integral Test and p-Series . . . . . . . . . . . . . . . . 260
Section 9.4
Comparisons of Series . . . . . . . . . . . . . . . . . . . . 270
Section 9.5
Alternating Series . . . . . . . . . . . . . . . . . . . . . . 277
Section 9.6
The Ratio and Root Tests . . . . . . . . . . . . . . . . . . 286
Section 9.7
Taylor Polynomials and Approximations . . . . . . . . . . 298
Section 9.8
Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 308
Section 9.9
Representation of Functions by Power Series
Section 9.10 Taylor and Maclaurin Series
. . . . . . . 321
. . . . . . . . . . . . . . . . 329
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343
Problem Solving
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354
C H A P T E R
Infinite Series
Section 9.1
Sequences
2. an 1. an 2n
a1 21 2
a2 4
22
a3 23 8
a4 24 16
a5 25 32
32
4. an a1 a2 n
2
3
21
3n
n!
3. an 11
1
12
5
a4 34 81 27
4! 24
8
a4 a5 35 243 81
5! 120 40
a5 n
2
1
2
3
1
2
5
1
2
8. an 1n1
a1 2n
2
2
1
16 1
a3 32
9
2
a3 3
1
110
a4 42
16
2
1
a4 4
2
1
1
52
25
21
a3 2
a2 1
2
a5 4
33 27 9
3!
6
2
13
1
a2 22
4
15
21
a3 32
243
a1 3
a2 a5 sin
1nn12
n2
21
32 9
2! 2
a4 sin 2 0
7. an 2
a2 a3 sin
16
a4 81
21
a1 a2 sin 0
8
a3 27
1
3
3
1!
5. an sin
a5 2
5
n
21
a1 a1 sin
4
9
a5 9
2n
n3
a1 2 1
4 2
a2 4
5
a3 6
1
6
a4 8
7
a5 10 5
8
4
1
8
1
16
6. an 9. an 5 1
4
1
2
1
32
1
1
n n2
a1 5 1 1 5
a2 5 1 1 19
2 4
4
a3 5 1 1 43
3 9
9
a4 5 1
1
77
4 16 16
a5 5 1
121
1
5 25
25
233
234
Chapter 9
10. an 10 Infinite Series
6
2
n n2
12. a1 4, ak1 11. a1 3, ak1 2ak 1
a2 2a1 1
a1 10 2 6 18
23 1 4
3 25
a2 10 1 2
2
a3 2a2 1
24 1 6
2 2 34
a3 10 3 3
3
a4 2a3 1
1 3 87
a4 10 2 8
8
6
266
2
a5 10 5 25
25
1
13. a1 32, ak1 ak
2
1
1
a3 a2 16 8
2
2
1
1
a3 a22 122 48
3
3
1
1
a4 a3 8 4
2
2
1
1
a4 a32 482 768
3
3
1
1
a5 a4 4 2
2
2
1
1
a5 a42 7682 196,608
3
3
17. an 40.5n1, a1 4, a2 2,
decreases to 0; matches (e).
19. an1n, a1 1, a2 1, a3 1, etc.; matches (d).
22.
8
20. an 6
a4 3 2 1a
12
a5 4 2 1a
30
2
3
4
4n
, a 4, a2 8,
n! 1
eventually approaches 0; matches (b).
18. an 23.
12
18
−1
12
12
−1
−3
2
an n, n 1, . . . , 10
3
24.
2 2 1a
1n
1
1
, a1 1, a2 , a3 , etc.; matches (c).
n
2
3
4
−1
−1
a3 8
8
, a 4, a2 ,
n1 1
3
decreases to 0; matches (f).
1
1
a2 a12 62 12
3
3
21.
4
1
15. an 1
1
a2 a1 32 16
2
2
8n
16
, a 4, a2 ,
n1 1
3
increases towards 8; matches (a).
1 2 1a
210 1 18
1
14. a1 6, ak1 ak2
3
16. an k
a2 26 1 10
a5 2a4 1
k 2 1a
4
an 2 , n 1, . . . , 10
n
25. an 3n 1
3
a5 35 1 14
−1
12
Add 3 to preceding term.
−1
an a6 36 1 17
2n
, n 1, 2, . . . , 10
n1
− 10
an 160.5n1, n 1, . . . , 10
26. an n6
2
a5 5 6 11
2
2
a6 66
6
2
Section 9.1
1
28. an an1, a1 1
2
27. an1 2an, a1 5
a5 240 80
1
1
,a 16 6
32
a5 a6 280 160
29. an Sequences
235
3
2n1
a5 3
3
24 16
a6 3
3
25
32
Multiply the preceding term by 12.
3
30. an an1, a1 1
2
a5 33.
31.
10! 8!910
8!
8!
32.
25! 23!2425
23!
23!
910 90
81
243
,a 16 6
32
n 1! n!n 1
n1
n!
n!
34.
2425 600
n 2! n!n 1n 2
n!
n!
35.
2n 1!
2n 1!
2n 1! 2n 1!2n2n 1
n 1n 2
36.
2n 2! 2n!2n 12n 2
2n!
2n!
37. lim
n→ 5n2
5
2
38. lim 5 n2
n→ 1
2n2n 1
1
505
n2
2n 12n 2
39. lim
n→ 2n
2
2
lim
2
n→ 1 1n2
1
n2 1
41. lim sin
n→ 43.
1n 0
40. lim
n→ 42. lim cos
n→ 44.
3
−1
n→ 5
1 4n2
5
5
1
2n 1
2
−1
12
−1
The graph seems to indicate that the sequence converges
to 1. Analytically,
lim an lim
45.
lim
−1
12
n→ 5n
n2 4
n→ The graph seems to indicate that the sequence converges
to 0. Analytically,
n1
x1
lim
lim 1 1.
x→ x→ n
x
lim an lim
n→ 46.
2
−1
n→ 1
1
lim
0.
n32 x→ x32
4
12
−1
−2
12
−1
The graph seems to indicate that the sequence diverges.
Analytically, the sequence is
an 0, 1, 0, 1, 0, 1, . . ..
Hence, lim an does not exist.
n→ The graph seems to indicate that the sequence converges
to 3. Analytically,
lim an lim 3 n→ n→ 1
3 0 3.
2n
236
Chapter 9
47. lim 1n
n→ Infinite Series
n n 1
48. lim 1 1n
3 n
3 n
n→ 1
n→ 3n2 n 4 3
, converges
2n2 1
2
does not exist, (alternates between
0 and 2), diverges.
does not exist (oscillates between
1 and 1), diverges.
50. lim
49. lim
n→ 51. an 1, converges
13
5.
. . 2n 1
1
2n
3
5
2nn
2n 2n .
. . 2n 1 < 1
2n
2n
Thus, lim an 0, converges.
n→ 52. The sequence diverges. To prove this analytically, we use mathematical induction to show
n1
0
k1
that an ≥ 32 . Clearly, a1 1 ≥ 32 1. Assume that ak ≥ 32 . Then,
ak1 13
5.
. . 2k 12k 1
2k 1
3
ak
≥
k 1!
k1
2
2kk 11.
k1
Since
2k 1 3
≥ ,
k1
2
we have ak1 ≥
32 k, which shows that an
≥
32 n1 for all n.
1 1n
0, converges
n→ n
54. lim
1 1n
0, converges
n→ n2
lnn3
3 lnn
lim
n→ 2
2n
n
56. lim
53. lim
55. lim
n→ lim
n→ n→ 3 1
0, converges
2 n
lim
n→ n→ 34
n
1
0, converges
2n
(L’Hôpital’s Rule)
(L’Hôpital’s Rule)
57. lim
lnn
12 ln n
lim
n→ n
n
58. lim 0.5n 0, converges
0, converges
59. lim
n→ n→ n 1!
lim n 1
n→ n!
, diverges
60. lim
n→ n 2!
1
lim
0, converges
n→ nn 1
n!
n n 1 n n 1 lim n nn1 1 n
2
61. lim
n→ lim
n→ 2nn 1 2nn 1 lim 4n2n 1 21,
2
62. lim
n→ converges
2
2
n→ 2
np
0, converges
n→ en
63. lim
p > 0, n ≥ 2
2
n→ 1 2n
0, converges
n2 n
Section 9.1
64. an n sin
Sequences
237
1
n
1
Let f x x sin .
x
lim x sin
x→ 1x 2 cos1x
1
sin1x
1
lim
lim
lim cos cos 0 1
x→ x→ x x→ 1x
1x 2
x
(L’Hôpital’s Rule)
or,
lim
x→ sin1x
sin y
1
lim
1. Therefore lim n sin 1.
y →0
n→ 1x
y
n
k
n
k
n
65. an 1 lim 1 n→ n
n
66. lim 21n 20 1, converges
n→ lim 1 u1u
k ek
u→0
where u kn, converges
67. lim
n→ sin n
1
lim sin n 0,
n→ n
n
68. lim
n→ cos n
0, converges
n2
69. an 3n 2
converges (because sin n is
bounded)
70. an 4n 1
74. an 1n1
2n2
72. an 71. an n2 2
75. an 1 1 n1
n
n
76. an 1 78. an 1
n!
79. an 81. an 2n!, n 1, 2, 3, . . .
1
3
1n1
n2
2n 1
2n
73. an n1
n2
77. an n
n 1n 2
2n1 1
2n
1n1
5 . . . 2n 1
80. an xn1
n 1!
1n1 2nn!
2n!
82. an 2n 1!, n 1, 2, 3, . . .
83. an 4 1
1
< 4
an1,
n
n1
monotonic; an < 4, bounded
84. Let f x 3x
6
. Then f x .
x2
x 22
Thus, f is increasing which implies an is increasing.
an < 3, bounded
238
85.
Chapter 9
Infinite Series
n ? n1
≥
2n2 2n1 2
?
2n3n ≥ 2n2n 1
?
2n ≥ n 1
86. an nen2
a1 0.6065
a2 0.7358
a3 0.6694
n ≥ 1
Not monotonic; an ≤ 0.7358, bounded
n ≥ 1
Hence,
2n ≥ n 1
2n3n
≥ 2n2n 1
n
n1
≥ n1 2
2n2
2
an ≥ an1.
1
True; monotonic; an ≤ 8, bounded
87. an 1n
1n
32
88. an a1 1
a2 a1 1
2
a3 a2 1
3
Not monotonic; an ≤ 1, bounded
90. an 32 < 32
n
n1
an1
Monotonic; lim an , not
n→ bounded
n
89. an a3 8
27
Not monotonic; an ≤ 23, bounded
91. an sin
n6
92. an cos
a1 0.500
a1 cos
a2 0.8660
a3 cos
94. an a3 0.3230
0
2
32 0
Not monotonic; an ≤ 1, bounded
sinn
n
a1 sin1
0.8415
1
a4 sin4
0.1892
4
a2 sin2
0.4546
2
a5 sin5
0.1918
5
a3 sin3
0.0470
3
a6 sin6
0.0466
6
a4 0.1634
Not monotonic; an ≤ 1, bounded
n2
a2 cos 1
a2 0.2081
an1
4
9
Not monotonic; an ≤ 1, bounded
a1 0.5403
n1
a4 0.8660
cos n
n
n
Monotonic; an ≤ 23, bounded
2
3
a3 1.000
93. an 23 > 23
Not monotonic, an
≤ 1, bounded
Section 9.1
1
n
95. (a) an 5 5
4
1
1
> 5
n
n1
an 5 Therefore, an converges.
(b)
3
< 4 ⇒ bounded
n
3
4
< 3
an1 ⇒ monotonic
n
n1
an 4 an1 ⇒ an, monotonic
239
3
n
96. (a) an 4 1
≤ 6 ⇒ an, bounded
n
Sequences
Therefore, an converges.
(b)
8
7
−1
−1
−1
lim 5 n→ 97. (a) an an 1
1
1 n
3
3
n→ 1
5
n
98. (a) an 4 1
⇒ an,bounded
3
<
4
< 131 3 1 1
1
> 4 n1
2n
2
Therefore, an converges.
(b)
0.4
12
6
−1
−0.1
lim
1
2n
an1 ⇒ an, monotonic
Therefore, an converges.
−1
3
4
n
1
≤ 4.5 ⇒ an, bounded
2n
an 4 n1
an1 ⇒ an, monotonic
n→ lim 4 1
1
1 n
3
3
1
1
1 n
3
3
(b)
12
0
12
12
−1
131 31 31
lim 4 n
n→ 99. an has a limit because it is a bounded, monotonic
sequence. The limit is less than or equal to 4, and greater
than or equal to 2.
2 ≤ lim an ≤ 4
1
4
2n
100. 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
(example: an 1n) or diverge (example: an n).
n→ 101. An P 1 r
12
n
(a) No, the sequence An diverges, lim An .
n→ The amount will grow arbitrarily large over time.
(b) P 9000, r 0.055, An 9000 1 0.055
12
A1 9041.25
A6 9250.35
A2 9082.69
A7 9292.75
A3 9124.32
A8 9335.34
A4 9166.14
A9 9378.13
A5 9208.15
A10 9421.11
n
240
Chapter 9
Infinite Series
102. (a) An 1004011.0025n 1
103. (a) A sequence is a function whose domain is the set of
positive integers.
A0 0
(b) A sequence converges if it has a limit. See the
definition.
A1 100.25
A2 200.75
(c) A sequence is monotonic if its terms are nondecreasing, or nonincreasing.
A3 301.50
A4 402.51
(d) A sequence is bounded if it is bounded below
an ≥ N for some N and bounded above an ≤ M
for some M.
A5 503.76
A6 605.27
(b) A60 6480.83
(c) A240 32,912.28
104. The first sequence because every
other point is below the x-axis.
107. an 105. an 10 3n
4n 1
1
n
106. Impossible. The sequence
converges by Theorem 9.5.
108. Impossible. An unbounded sequence diverges.
109. (a) An 0.8n 2.5 billion
(b) A1 $2 billion
110. Pn 16,0001.045n
P1 $16,720.00
A2 $1.6 billion
P2 $17,472.40
A3 $1.28 billion
P3 $18,258.66
A4 $1.024 billion
P4 $19,080.30
(c) lim 0.8 2.5 0
P5 $19,938.91
n
n→ 111. (a) an 6.60n2 151.7n 387
112. (a) an 244.2n 3250
7000
1500
5
500
(b) For 2008, n 18 and a18 7646 million.
10n
n!
114. an 1 (a) a9 a10 109
9!
1,000,000,000
362,880
13
0
(b) In 2008, n 18 and a18 979 endangered species.
113. an 2
13
1,562,500
567
1
n
n
a1 2.0000
a2 2.2500
a3 2.3704
a4 2.4414
a5 2.4883
(b) Decreasing
a6 2.5216
(c) Factorials increase more rapidly than exponentials.
lim 1 n→ 1
n
n
e
Section 9.1
n 115. an n n1n
Sequences
241
116. Since
lim sn L > 0,
a1 111 1
n→ there exists for each > 0,
a2 2 1.4142
an integer N such that sn L < for every n > N.
Let L > 0 and we have,
3 3 1.4422
a3 4
a4 4 1.4142
sn L
5 5 1.3797
a5 < L, L < sn L < L, or 0 < sn < 2L
for each n > N.
6 6 1.3480
a6 Let y lim n1n.
n→ ln y lim
n→ 0
1n ln n lim lnnn lim 1n
n
n→ n→ Since ln y 0, we have y e 1. Therefore,
n n 1.
lim 0
n→ 117. True
118. True
119. True
120. True
121. an2 an an1
(a) a1 1
a7 8 5 13
a2 1
a8 13 8 21
a3 1 1 2
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
a
(b) bn n1, n ≥ 1
an
b6 13
8
b2 2
2
1
b7 21
13
3
b3 2
34
b8 21
5
b4 3
55
b9 34
122. x0 1, xn 1
an1
an
an an1 an1
bn
an
an
(d) If lim bn , then lim 1 n→ n→ n→ 1
1
1
8
5
1
1
1
bn1
anan1
1
.
bn1
Since lim bn lim bn1, we have
b1 b5 (c) 1 b10 1 1 .
n→ 1 2
0 2 1
1 ± 1 4 1 ± 5
2
2
Since an, and thus bn, is positive, 1 5
1.6180.
2
89
55
1
1
, n 1, 2, . . .
x
2 n1 xn1
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.
242
Chapter 9
Infinite Series
123. (a) a1 2 1.4142
(b) an 2 an1,
n ≥ 2, a1 2
a2 2 2 1.8478
a3 2 2 2 1.9616
a4 2 2 2 2 1.9904
a5 2 2 2 2 2 1.9976
(c) We first use mathematical induction to show that an ≤ 2; clearly a1 ≤ 2. So assume ak ≤ 2. Then
ak 2 ≤ 4
ak 2 ≤ 2
ak1 ≤ 2.
Now we show that an is an increasing sequence. Since an ≥ 0 and an ≤ 2,
an 2an 1 ≤ 0
an2 an 2 ≤ 0
an2 ≤ an 2
an ≤ an 2
an ≤ an1.
Since an is a bounding increasing sequence, it converges to some number L, by Theorem 9.5.
lim an L ⇒ 2 L L ⇒ 2 L L2 ⇒ L2 L 2 0
n→ ⇒ L 2L 1 0 ⇒ L 2 L 1
124. (a) a1 6 2.4495
(b) an 6 an1,
n ≥ 2, a1 6
a2 6 6 2.9068
a3 6 6 6 2.9844
a4 6 6 6 6 2.9974
a5 6 6 6 6 6 2.9996
(c) We first use mathematical induction to show that an ≤ 3; clearly a1 ≤ 3. So assume ak ≤ 3. Then
6 ak ≤ 9
6 ak ≤ 3
ak1 ≤ 3.
Now we show that an is an increasing sequence. Since an ≥ 0 and an ≤ 3,
an 3an 2 ≤ 0
an2 an 6 ≤ 0
an2 ≤ an 6
an ≤ an 6
an ≤ an1.
Since an is a bounded increasing sequence, it converges to some number L: lim an L. Thus,
6 L L ⇒ 6 L L2 ⇒ L2 L 6 0
⇒ L 3L 2 0 ⇒ L 3 L 2
n→ Section 9.1
1 1 4k
.
2
(c) lim an L implies that
n→ [Note that if k 2, then an ≤ 3, and if k 6, then
an ≤ 3.] Clearly,
1 4k
a1 k ≤
2
≤
L k L ⇒ L2 k L
⇒ L2 L k 0
1 1 4k
.
2
⇒ L
Before proceeding to the induction step, note that
Since L > 0, L 2 21 4k 4k 2 21 4k 4k
1 ± 1 4k
.
2
1 1 4k
.
2
1 1 4k
1 21 4k 1 4k
k
2
4
1 1 4k
1 1 4k
k
2
2
1 1 4k
2
So assume an ≤
an k ≤
an k ≤
an1 ≤
k
2
1 1 4k
.
2
1 1 4k
. Then
2
1 1 4k
k
2
1 1 4k
2
k
1 1 4k
.
2
an is increasing because
a 1 n
1 4k
2
a 1 n
1 4k
2
≤0
an2 an k ≤ 0
an2 ≤ an k
an ≤ an k
an ≤ an1.
126. (a) a0 10, b0 3
a1 a0 b0 10 3
6.5
2
2
b1 a0b0 103 5.4772
a2 a1 b1
5.9886
2
b2 a1b1 5.9667
a3 a2 b2
5.9777
2
b3 a2b2 5.9777
a4 a3 b3
5.9777
2
b4 a3b3 5.9777
a5 a4 b4
5.9777
2
b5 a4b4 5.9777
The terms of an are decreasing, and those of bn are increasing. They both seem to approach the same limit.
—CONTINUED—
243
(b) Since an is bounded and increasing, it has a limit L.
125. (a) We use mathematical induction to show that
an ≤
Sequences
244
Chapter 9
Infinite Series
126. —CONTINUED—
(b) For n 0, we need to show a0 > a1 > b1 > b0. Since a0 > b0, 2a0 > a0 b0 ⇒ a0 >
a0 b0
a 1.
2
Since a0 b02 > 0,
a02 2a0b0 b02 > 0 ⇒ a02 2a0b0 b02 > 4a0b0
⇒ a0 b02 > 4a0b0 ⇒ a0 b0 > 2a0b0 ⇒ a1 > b1.
Since a0 > b0, a0b0 > b02 ⇒ a0b0 > b0 ⇒ b1 > b0. Thus, we have shown that a0 > a1 > b1 > b0.
Now assume ak > ak1 > bk1 > bk. Since ak1 > bk1,
2ak1 > ak1 bk1 ⇒ ak1 >
ak1 bk1
ak2.
2
Since ak1 bk12 > 0,
ak12 2ak1bk1 bk12 > 0 ⇒ ak12 2ak1bk1 bk12 > 4ak1bk1
⇒ ak1 bk12 > 4ak1bk1
⇒ ak1 bk1 > 2ak1bk1
⇒ ak2 > bk2.
Since ak1 > bk1, ak1bk1 > bk12 ⇒ ak1bk1 > bk1 ⇒ bk2 > bk1.
Thus, we have shown that ak1 > ak2 > bk2 > bk1.
(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→ 127. (a) f x sin x, an n sin
AB
⇒ 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
f 0 h f 0
h
lim
f h
h
lim
f 1n
1n
h→0
n→ lim n f
n→ lim an
n→ 1
sin1n
lim
1 f 0
n n→ 1n
1n
128. an nr n
1
1
(a) r : an n
2
2
n
n
→0
2n
(b) r 1: an n, diverges
3
3 2
, diverges
(c) r : an n
2
2
(d) If r ≥ 1, diverges. If r < 1,
nr n →
n
rn
“”
1
rn
→ 0.
rn lnr lnr
The sequence converges for r < 1.
Section 9.1
129. (a)
y
(b)
y = ln x
2.0
y = lnx
2.0
1.5
1.0
1.0
0.5
2
3
4
x
...
n
2
n
ln x dx < ln 2 ln 3 . . . ln n
1
(c)
y
1.5
0.5
3
...
4
x
n+1
n1
ln x dx > ln 2 ln 3 . . . ln n
1
ln1
23.
. . n
lnn!
lnn!
ln x dx x ln x x C
n
ln x dx n ln n n 1 ln nn n 1
1
From part (a): ln nn n 1 < lnn!
n n1
eln n
< n!
nn
< n!
en1
n1
ln x dx n 1 lnn 1 n 1 1 lnn 1n1 n
1
From part (b): lnn 1n1 n > lnn!
n 1 n
elnn1
> n!
n 1n1
> n!
en
(d)
Sequences
nn
en1
n
e1 1n
< n! <
n 1n1
en
n n! <
<
(e) n 20:
n 1n1n
e
n 50:
n n!
1
n 11 1n
<
<
e1 1n
n
ne
lim
n→ lim
n→ n 100:
1
1
e1 1n
e
20
20!
20
50
50!
50
100
1
0.3679
e
n 11 1n
n 1 n 11n
lim
n→ ne
n
e
1 1
1 e
e
By the Squeeze Theorem, lim
n→ n
n!
n
1
.
e
0.4152
0.3897
100!
0.3799
100
245
246
Chapter 9
130. an Infinite Series
1 n
1
n k1 1 kn
(a) a1 1
1
0.5
11 2
a2 1
1
1 2 1
7
1
0.5833
2 1 12 1 1
2 3 2
12
a3 1
1
1
37
1
0.6167
3 1 13 1 23 1 1
60
a4 1
1
1
1
533
1
0.6345
4 1 14 1 24 1 34 1 1
840
a5 1
1
1
1
1
1
1627
0.6456
5 1 15 1 25 1 35 1 45 1 1
2520
1 n
1
n→ n k1 1 kn
(b) lim an lim
n→ 1
0
131. For a given > 0, we must find M > 0 such that
1
< n3
an L whenever n > M. That is,
n3 >
1
1
or n >
13
.
So, let > 0 be given. Let M be an integer satisfying
M > 113. For n > M, we have
n >
n3 >
>
1
n→ 1
132. For a given > 0, we must find M > 0 such that
an L r n whenever n > M. That is,
n ln r < ln or
n >
ln
(since ln r < 0 for r < 1).
ln r
So, let > 0 be given. Let M be an integer satisfying
M >
ln
.
ln r
n >
1
1
⇒ 3 0 < .
n3
n
Thus, lim
For n > M, we have
13
1
0 ln 2
1
dx ln 1 x
1x
1
0.
n3
ln
ln r
lnrn < ln
rn < r n 0 < .
n ln r < ln
Thus, lim r n 0 for 1 < r < 1.
n→ 133. If an is bounded, monotonic and nonincreasing,
then a1 ≥ a2 ≥ a3 ≥ . . . ≥ an ≥ . . . . Then
a1 ≤ a2 ≤ a3 ≤ . . . ≤ an ≤ . . . is a
bounded, monotonic, nondecreasing sequence which
converges by the first half of the theorem. Since an
converges, then so does an.
134. Define an xn1 xn1
,
xn
n ≥ 1.
xn12 xn xn2 1 xn2 xn1xn1 ⇒
xn1xn1 xn1 xnxn xn2
xn1 xn1 xn2 xn
xn
xn1
an an1
Hence, a1 a2 . . . a. Thus,
xn1 an xn xn1 axn xn1.
Section 9.2
Series and Convergence
135. Tn n! 2n
We use mathematical induction to verify the formula.
T0 1 1 2
T1 1 2 3
T2 2 4 6
Assume Tk k! 2k. Then
Tk1 k 1 4Tk 4k 1Tk1 4k 1 8Tk2
k 5k! 2k 4k 1k 1! 2k1 4k 4k 2! 2k2
k 5kk 1 4k 1k 1 4k 1k 2! k 54 8k 1 4k 12k2
k2 5k 4k 4 4k 1! 8 2k2
k 1! 2k1.
By mathematical induction, the formula is valid for all n.
Section 9.2
Series and Convergence
2. S1 16 0.1667
1. S1 1
S2 1 14 1.2500
S2 16 16 0.3333
S3 1 14 19 1.3611
3
S3 16 16 20
0.4833
1
S4 1 14 19 16
1.4236
3
2
S4 16 16 20
15
0.6167
1
1
S5 1 14 19 16
25
1.4636
1
1
3
2
5
S5 6 6 20 15 42 0.7357
4. S1 1
3. S1 3
S2 3 1.5
S2 1 13 1.3333
S3 3 92 27
4 5.25
S3 1 13 15 1.5333
81
S4 3 92 27
4 8 4.875
S4 1 13 15 17 1.6762
9
27
81
243
S5 3 2 4 8 16 10.3125
1
1
1
1
S5 1 3 5 7 9 1.7873
9
2
6. S1 1
5. S1 3
7.
S2 3 32 4.5
S2 1 12 0.5
S3 3 32 34 5.250
S3 1 12 16 0.6667
S4 3 32 34 38 5.625
1
S4 1 12 16 24
0.6250
3
3
3
3
S5 3 2 4 8 16 5.8125
1
1
S5 1 12 16 24
120
0.6333
3
2
n
3
8.
4
3
n
9.
10001.055
n
n0
n0
n0
Geometric series
Geometric series
Geometric series
r
3
> 1
2
Diverges by Theorem 9.6
r
4
> 1
3
Diverges by Theorem 9.6
r 1.055 > 1
Diverges by Theorem 9.6
247
248
10.
Chapter 9
Infinite Series
21.03
n
11.
n0
lim
n→ > 1
n
2
n1
n
10
n1
n2
1
Diverges by Theorem 9.9
2n 1
n1
n1 2
9 1
n0 4 4
n
1
n
n0
4 4 15
1
n
9
5 45
,S 4 16 2 4
S0 n
3 2 17
1
n
3
S0 17
1
n0
n
154
154
3
1 14
54
17
1 1
1 . . .
3
2 4
1
1
3
1 23 13
17
8
n
17
, S 0.63, S3 5.1, . . .
3 1
17
8
3
9
n0
5
22.
2
n
n
1
n0
173
173
3
1 89 179
4
2
. . .
5 25
7
S0 1, S1 , . . .
5
Matches (e). The series is geometric:
5
2
n
n0
1
5
1 25 3
nn 1 n n 1 1 2 2 3 3 4 4 5 . . .,
n1
1
1
1
1
1
n1
1
1
lim Sn lim 1 1
n→
n→
n
n
1
n
1
n1
17
8 64 . . .
1 3
9 81
Matches (d). Analytically, the series is geometric:
1
17
17 2 34
3 1 12
3 3
9
S0 2 4 . . .
3 9
n0
Matches (f). The series is geometric:
n
3 9 20.
17
17
,S , . . .
3 1
6
3 2 2
n0
n0
1
Matches graph (b). Analytically, the series is geometric:
15
1
1
. . .
1 4
4 16
n
5
S0 1, S1 , S2 2.11, . . . .
3
Matches graph (a). Analytically, the series is geometric:
15 1
4
n0 4
2
n0
21
2.95, . . .
16
15
45
, S , S 3.05, . . .
4 1 16 2
3
18.
94
94
3
1 14 34
n0
10
n!
2n
lim
1
1 1n2
Diverges by Theorem 9.9
n→ n
n→ 9
1
1
1 . . .
4
4 16
lim
n!
n1
44
9
n2 1
2
16.
2n 1
1 2n 1
0
lim
n1
n→
2
2
2
n
lim
Matches graph (c). Analytically, the series is geometric:
23.
n
1
2
Diverges by Theorem 9.9
9
9
S0 , S1 4
4
21.
n1
1
n
0
2n 3 2
Diverges by Theorem 9.9
n
14.
Diverges by Theorem 9.9
19.
lim
n→ n→ lim
n
n1
2
n→ 17.
2n 3
n2
10
n→ n 1
lim
15.
12.
Diverges by Theorem 9.9
Diverges by Theorem 9.6
13.
n
n1
Geometric series
r 1.03
n1
1
1
1
1
1
Sn 1 1
n1
Section 9.2
24.
Series and Convergence
nn 2 2n 2n 2 2 6 4 8 6 10 8 12 10 14 . . .
1
n1
1
1
1
1
1
1
1
1
1
1
1
1
n1
1 1
1
1
1
1 1 3
lim Sn lim
n→
n→
n
n
2
2
4
2
n
1
2
n
2
2
4 4
n1
25.
2 4 3
n
26.
3
4
Geometric series with r n
Geometric series with r 12 < 1
< 1
Converges by Theorem 9.6
Converges by Theorem 9.6
1
n0
n0
27.
2 2 0.9
n
28.
0.6
n
n0
n0
Geometric series with r 0.9 < 1
Geometric series with r 0.6 < 1
Converges by Theorem 9.6
Converges by Theorem 9.6
29. (a)
nn 3 2 n n 3
6
n1
1
1
n1
1 41 12 51 13 61 14 71 . . .
2
S
n
1 1
1
1
1
2 3
n1 n2 n3
2 1
2 1
(b)
1 1
11
3.667
2 3
3
n
5
10
20
50
100
Sn
2.7976
3.1643
3.3936
3.5513
3.6078
(c)
(d) The terms of the series decrease in magnitude slowly. Thus,
the sequence of partial sums approaches the sum slowly.
30. (a)
0
11
0
nn 4 n n 4
4
n1
1
1
n1
1
1
(b)
5
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
2.0833
2 3 4 12
n
5
10
20
50
100
Sn
1.5377
1.7607
1.9051
2.0071
2.0443
(d) The terms of the series decrease in magnitude slowly. Thus,
the sequence of partial sums approaches the sum slowly.
(c)
4
0
11
0
249
250
Chapter 9
31. (a)
Infinite Series
20.9
n1
n1
(b)
20.9
n
n0
2
20
1 0.9
(c)
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. Thus,
the sequence of partial sums approaches the sum slowly.
32. (a)
30.85
n1
n1
(b)
3
20
1 0.85
(Geometric series)
(c)
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. Thus,
the sequence of partial sums approaches the sum slowly.
33. (a)
100.25
n1
n1
(b)
10
40
13.3333
1 0.25
3
(c)
n
5
10
20
50
100
Sn
13.3203
13.3333
13.3333
13.3333
13.3333
15
11
0
0
(d) The terms of the series decrease in magnitude rapidly. Thus,
the sequence of partial sums approaches the sum rapidly.
34. (a)
5 3 1
n1
n1
(b)
5
15
3.75
1 13
4
(c)
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. Thus,
the sequence of partial sums approaches the sum rapidly.
35.
n
n2
36.
2
12
1
12
1
1
1 1 n2 n 1 n 1
2 n2 n 1 n 1
1
2
1
3
1
1
2
2
4
1
1
1
1
1
1
nn 2 2 n n 2 2 1 3 2 4 3 5 . . . 21 2 3
4
1
1
1
1
1
1
1
1
n1
n 1n 2 8 n 1 n 2 8 2 3 3 4 4 5 . . . 82 4
8
n1
38.
1 31 2 4 3 5 4 6 . . .
n1
37.
1
1
1
1
1
1
1
1
1
n1
2n 12n 3 2 2n 1 2n 3 2 3 5 5 7 7 9 . . . 23 6
n1
1
1
n1
1
1
1
1
1
1
1
1
1
1 1
1
Section 9.2
39.
1
2
n
1
2
1 12
n0
6 5 40.
4
n
n0
Series and Convergence
6
30
1 45
41.
1
10
1 110
9
44.
2 1
n
251
1
2
1 12 3
8
32
1 34
n0
(Geometric)
42.
2 3 n
n
2
n0
45.
1
3
3 n0
47.
2
6
1 23 5
1
1
1
2 3 2
n0
10
1
n
n0
3
9
1 13 4
n
43.
n
n
n0
46.
1
3
n
48.
1
2
n0
4 n
0.7n 0.9n n1
sin1
n
7
10
n
9
10
n0
3 1
2 2
49. Note that sin1 0.8415 < 1. The series
n
4
8
1 12 3
1
1
1 12 1 13
2
3
n0
n0
8 4 n
2
n0
1
1
2
1 710 1 910
34
10
10 2 3
3
is geometric with r sin1 < 1. Thus,
n1
sin1
n
sin1
n1
50. Sn n
n0
n
k1
sin1
9k2
sin1
5.3080.
1 sin1
n
1
1
3k 2 k1 3k 13k 2
9k 3 9k 6 3 3k 1 3k 2
n
1
1
1
k1
n
1
1
k1
12 51 15 81 18 111 . . . 3n 1 1 3n 1 2
1
3
1 1
1
3 2 3n 2
lim Sn lim
n→ n→ 51. (a) 0.4 1 1
1
1
3 2 3n 2
6
1010
4
1
n
52. (a) 0.9 n0
4
1
(b) Geometric series with a 10 and r 10
S
81 1
100
100
n0
S
(b) 0.9 n
(b) Geometric series with a 910
1
n
n1
410
4
a
1 r 1 110 9
53. (a) 0.81 9
9
9
. . .
10 100 1000
100
1
n
n1
and r a
81100
81
9
1 r 1 1100 99 11
1
100
n
9
1
1
10 1 110
54. (a) 0.01 81
100
9 1
10 n0 10
(b) 0.01 1
100
1
1 1
100 n0 100
1
n
1 1100 100 1
100
99
99
252
Chapter 9
Infinite Series
40100
55. (a) 0.075 3
1
n
56. (a) 0.215 n0
3
1
(b) Geometric series with a 40 and r 100
S
57.
340
5
a
1 r 99100 66
S
58.
n 10
1
0
10n 1 10
n1
n1
1
0
2n 1 2
lim
Diverges by Theorem 9.9
n n 2 1 3 2 4 3 5 4 6 . . .
1
1
a
1
3200
71
5 1 r 5 99100 330
2n 1
n→ Diverges by Theorem 9.9
59.
1
1
1
1
1
1
1
1
1
n1
60.
1 3
, converges
2 2
nn 3 3 n n 3
1
1
n1
61.
1
n1
n1
lim
n
3
10n 1
n→ (b) Geometric series with a 200 and r 100
n 10
3
1
1
5 n0 200 100
1
1
n1
1 41 12 51 13 61 14 71 15 81 16 91 . . .
1
3
1
1 1
11
1 , converges
3
2 3
18
3n 1
2n 1
62.
n1
lim
n→ 3n
n
3
n1
3n 1 3
0
2n 1 2
3n
ln 23n
lim
3
2
n→
n
3n
lim
n→ ln 22 3n
ln n3 3n
lim
n→ n→ 6n
6
Diverges by Theorem 9.9
lim
(by L’Hôpital’s Rule); diverges by Theorem 9.9
63.
4
2
n0
n
4
2
1
n
64.
n0
1
n0
1
65.
4
n
1
Geometric series with r 2
Geometric series with r 4
Converges by Theorem 9.6
Converges by Theorem 9.6
1.075
n
66.
2n
100
n0
n1
Geometric series with r 1.075
Geometric series with r 2
Diverges by Theorem 9.6
Diverges by Theorem 9.6
67. Since n > lnn, the terms an n
lnn
do not approach 0 as n → . Hence, the series
n
lnn diverges.
n2
68. Sn ln k lnk
n
k1
1
n
k1
0 ln 2 ln 3 . . . lnn
Since lim Sn diverges,
n→ lnn diverges.
n1
1
Section 9.2
69. For k 0,
e
70.
n
n1
k
lim 1 n→ n
n
lim
n→ k
1
n
e
1
n
Series and Convergence
71. lim arctan n converges since
n→ n1
it is geometric with
nk k
Hence,
1
r e < 1.
ek 0.
253
0
2
arctan n diverges.
n1
For k 0, lim 1 0n 1 0.
n→ 1 n
Hence,
k
n
diverges.
n1
ln
n
72. Sn k1
ln
k1
k
21 ln32 . . . lnn n 1
ln 2 ln 1 ln 3 ln 2 . . . lnn 1 ln n
lnn 1 ln1 lnn 1
Diverges
73. See definition on page 606.
74. lim an 5 means that the limit of the sequence an is 5.
n→ a
n
75. The series given by
ar
a1 a2 . . . 5 means that the limit of the
a ar ar 2 . . . ar n . . . , a 0
n
n0
n1
partial sums is 5.
is a geometric series with ratio r. When 0 < r < 1,
the series converges to a1 r. The series diverges if
r ≥ 1.
76. If lim an 0, then
n→ n→ n
diverges.
n1
n
lim an lim
n→ a
n
n1
an 77.
a
78. (a)
a
n
a1 a2 a3 . . .
k
a1 a2 a3 . . .
n1
n1
1
n
an converges
(b)
These are the same. The third series is different,
unless a1 a2 . . . a is constant.
diverges because lim an 0.
n→ n1
a
k1
(c)
a
k
ak ak . . .
n1
79.
xn
2
n1
n
2
x
n
n1
x x
2 n0 2
n
Geometric series: converges for
f x x x
2 n0 2
n
80.
n
3x
n1
x
< 1 or x < 2
2
x
x 2
1
x
,
2 1 x2 2 2 x 2 x
3x
n
n0
Geometric series: converges for 3x < 1 ⇒ x <
f x 3x
3x
n0
x < 2
3x
n
3x
1
3x
,
1 3x 1 3x
1
3
1
x < 3
254
81.
Chapter 9
x 1
n
Infinite Series
x 1
n1
x 1
n0
n0
Geometric series: converges for x 1 < 1 ⇒ 0 < x < 2
f x x 1
x3
4
1 x
n n
n
1
x1
,
1 x 1 2 x
f x 0 < x < 2
n0
4
n
n0
n
4
1 x 34
16
4
, 1 < x < 7
4 x 34 7 x
1 x
84.
x3
4
n 2n
n0
x 2 n
n0
Geometric series: converges for
Geometric series: converges for
x < 1
x 2 < 1
⇒ x < 1 ⇒ 1 < x < 1
f x x
n
n0
85.
n0
x
n
x3
< 1 ⇒ x 3 < 4 ⇒ 1 < x < 7
4
x 1
83.
Geometric series: converges for
n0
x 1
4
82.
n
x
1
1
,
1x
f x 1 < x < 1
n0
n1
1
< 1
Geometric series: converges if
x
x
n0
1
2 n
2
x2
4
n
x2
1
1
,
1 x2
1 x2
x2
x2
2
4 n0 x 4
Geometric series: converges for
⇒ x > 1 ⇒ x < 1 or x > 1
f x x
86.
n
x n0
n
⇒ 1 < x < 1
⇒
1
x
, x > 1 or x < 1
1 1x x 1
x2
f x <
x2
n
x2
< 1
x2 4
4 ⇒ converges for all x
x2
x2
1 < x < 1
4
1
1
x2
x2
x2 4
x2 x2 4 x2
4 4
4
88. Neither statement is true. The formula
87. (a) Yes, the new series will still diverge.
1
1 x x2 . . .
x1
(b) Yes, the new series will converge.
holds for 1 < x < 1.
(c) y1 89. (a) x is the common ratio.
(b) 1 x x 2 . . . xn n0
1
,
1x
x < 1
1
1x
3
f
y2 S3 1 x x 2
y3 S5 1 x x 2 x 3 x 4
S5
S3
−1.5
1.5
0
90. (a)
2x is the common ratio.
(c) y1 x
x2 x3
x
(b) 1 . . . 2
4
8
2
n0
n
1
1 x2
2
,
2x
x < 2
2
2x
5
f
x2
x
y2 S3 1 2
4
y3 S5 1 x
x2 x3
x4
2
4
8
16
S5
S3
−5
5
−5
Section 9.2
1 0.5x
1 0.5 91. f x 3
7
11 0.8
0.8 1
S
−2
2 5 10
S
1
< 0.0001
nn 1
2
10
1 45
10,000 < 2n
12
This inequality is true when n 14.
0.01n < 0.0001
4110,000
2
10,000 < 10n
Choosing the positive value for n we have n 99.5012.
The first term that is less than 0.0001 is n 100.
18
16
−2
1
< 0.0001
2n
94.
0 < n2 n 10,000
1 ±
−2
The horizontal asymptote is the sum of the series.
f n is the nth partial sum.
10,000 < n2 n
n
n
n0
The horizontal asymptote is the sum of the series.
f n is the nth partial sum.
93.
4
−1
3
6
1 12
y = 10
12
Horizontal asymptote: y 10
n
n0
255
x
92. f x 2
y=6
Horizontal asymptote: y 6
3 2 Series and Convergence
This inequality is true when n 5. This series converges
at a faster rate.
n
< 0.0001
10,000 < 8n
This inequality is true when n 5. This series converges
at a faster rate.
80001 0.9n1 1
1 0.9
n1
95.
80000.9 i
i0
96. Vt 225,0001 0.3n 0.7n225,000
V5 0.75225,000 $37,815.75
80,0001 0.9n, n > 0
1001 0.75n1 1
1 0.75
n1
97.
1000.75 i
i0
n1
98.
1000.60i i0
4001 0.75n million dollars
Sum 400 million dollars
100. The ball in Exercise 99 takes the following times for
each fall.
D2 0.8116 0.8116 320.81
down
D3 160.812 160.812 320.812
D 16 320.81 320.812 . . .
16 320.81
n0
152.42 feet
n
2501 0.6n million dollars
Sum 250 million dollars
99. D1 16
up
1001 0.6n
1 0.6
16 32
1 0.81
s1 16t 2 16
s1 0 if t 1
s2 16t 2 160.81
s2 0 if t 0.9
s3 16t 2 160.812
s3 0 if t 0.92
sn 16t 2
160.81
n1
sn 0 if t 0.9n1
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
t12
0.9
n1
1 n
1 2
0.9
n0
2
19 seconds.
1 0.9
n
256
Chapter 9
101. Pn P2 1 1
2 2
1 1
2 2
22
1 1
n
n0
Infinite Series
n
102. Pn 2
1
8
P2 1 2
3 3
33
12
1
1 12
1 2
3 3
1 2
n0
n
n
103. (a)
1
2
n
n1
2
4
27
n0
1 1
2 2
1
1
1
2 1 12
(b) No, the series is not geometric.
13
1
1 23
(c)
n 2 1
n
2
n1
1
1
1 1
1
4 7. . .
2 2
2
2 n0 8
104. Person 1:
n
1
1
1
1 1
. . .
22 25 28
4 n0 8
n
Person 2:
1
1
1
1 1
. . .
23 26 29
8 n0 8
n
Person 3:
Sum:
1
4
1
2 1 18 7
1
1
2
4 1 18 7
1
1
1
8 1 18 7
4 2 1
1
7 7 7
105. (a) 64 32 16 8 4 2 126 in.2
1
2
(b)
n
64
n0
64
128 in.2
1 12
16 in.
Note: This is one-half of the area of the original square!
16 in.
106. (a) sin Yy1
z
x1y1
Yy1
x1y2
sin x1y1
sin ⇒ Yy1 z sin (b) If z 1 and ⇒ x1y2 x1 y1 sin z sin3 Total: z sin z sin2 z sin3 . . . z
50,0001.06
20
107.
1
n
12
, then total 1.
6
1 12
⇒ x1y1 Yy1 sin z sin2 50,000 19 1
1.06 i0 1.06
50,000 1 1.0620
,
1.06 1 1.061
n1
sin 1 sin i
n 20, r 1.061
573,496.06
1
108. Surface area 4 12 94 3 2
109. w n1
0.012 i
i0
92 4 19 2 .
0.011 2n
0.012n 1
12
(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
n
. . 4 . . . Section 9.2
P1 12
12t1
110.
r
n
n0
12t1
12r 12
12
r
r
P 1 1 P 1 r
r
12
r
12
1 1 12
12t
n0
111. P 50, r 0.03, t 20
(b) A 1220
1 $16,415.10
(a) A 75
50 e0.0320 1
$16,421.83
e0.0312 1
(b) A 12
0.05
1 0.05
12 1225
1 $44,663.23
75e0.0525 1
$44,732.85
e0.0512 1
114. P 20, r 0.06, t 50
113. P 100, r 0.04, t 40
(b) A 1
112. P 75, r 0.05, t 25
12
0.03
1 0.03
12 (a) A 100
12t
P1 e r1212t Pe rt 1
r12
1 e r12
e
1
Pe r12n (a) A 50
257
12t
P 1 1
Series and Convergence
12
0.04
1 0.04
12 1240
(a) A 20
1 $118,196.13
100e0.0440 1
$118,393.43
e0.0412 1
(b) A 115. (a) an 3484.13631.0502n 3484.1363e0.04897n
12
0.06
1 0.06
12 1250
1 $75,743.82
20e0.0650 1
$76,151.45
e0.0612 1
(b) Adding the ten values a3, a4, . . . , a12 you obtain
12
a
6500
n
50,809, or $50,809,000,000.
n3
12
(c)
3484.1363e
0.04897n
50,803, or $50,803,000,000
n3
2
3500
13
(Answers will vary.)
116. T 40,000 40,0001.04 . . . 40,0001.0439
39
40,0001.04
n
40,000
n0
117. False. lim
n→ 11 1.04 $3,801,020
1.0440
118. True
119. False;
ar
n1
n
1 a r a
The formula requires that the geometric series begins with n 0.
121. True
122. True
0.74999 . . . 0.74 1
1
0, but
diverges.
n
n1 n
9
9
. . .
103 104
0.74 9 1
103 n0 10
0.74 9
103
1 110
0.74 9
103
0.74 1
0.75
100
1
10
9
n
120. True
lim
n→ n
1
0
1000n 1 1000
258
Chapter 9
Infinite Series
124. Let Sn be the sequence of partial sums for the
convergent series
123. By letting S0 0, we have
n
a
an k
n1
a
k
k1
n1
a
n
an Sn Sn1. Thus,
k1
Sn Sn1
Rn n1
S
c S
n1
ak L Sn,
we have
n1
n→ kn1
Sn1 c c
n
L. Then lim Sn L and since
n1
lim Rn lim L Sn
c Sn.
n→ n1
n→ lim L lim Sn L L 0.
n→ 125. Let
a
1 and b
n
n
n0
126. If an bn converged, then an bn an bn would converge, which is a contradiction. Thus,
an bn diverges.
1.
n0
Both are divergent series.
a
n
n0
n0
1 1 1 1 0
bn 127. Suppose, on the contrary, that can converges. Because
c 0,
c ca
1
n
a
n
converges. This is a contradiction since can diverged.
Hence, can diverges.
129. (a)
128. If an converges, then lim an 0. Hence,
n→ 1
lim
0,
n→ an
which implies that 1an diverges.
an3 an1
1
1
an1an2 an2an3 an1an2an3
n
k0
1
an1an3
k1ak3
n
1
1
a
a
k1 k2
k2ak3
k0
a 1a
1 2
1
1
1
1
1
. . .
a2a3
a2a3 a3a4
an1an2 an2an3
1
1
a1a2 an2an3
1
1
an2an3
1
1
lim Sn lim 1 1
n→
n→
a
a
n1 n3
n2an3
a
n0
an2
an1an2an3
a
1
a
(b) Sn n→ Section 9.2
Series and Convergence
259
130. S2n 1 2x x 2 2x3 . . . x2n 2x2n1
1 x 2 x 4 . . . x2n 2x 2x3 . . . 2x2n1
n
x 2 k
2x
k0
n
x 2 k
k0
As n → , the two geometric series converge for x < 1:
1
1 2x
1
lim S2n 2x
,
n→ 1 x2
1 x2
1 x2
131.
1 1
1
1
1
2 3. . .
r
r
r
n0 r r
n
132. The entire rectangle has area 2 because the height is 1
1
1
and the base is 1 2 4 . . . 2. The squares all
lie inside the rectangle, and the sum of their areas is
1r
1 1r
1
r1
1
since
< 1
r
This is a geometric series which converges if
x < 1
1
1
1
1
. . ..
22 32 42
Thus,
1
n
n1
2
< 2.
1
< 1 ⇔ r > 1.
r
133. 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 Pekt Pe2kt . . . Pen1kt where k ln 2H.
One time interval after the last dose is administered is given by Tn1 Pekt Pe2kt Pe3kt . . . Penkt.
Two time intervals after the last dose is administered is given by Tn2 Pe2kt Pe3kt Pe4kt . . . Pen1kt
and so on. Since k < 0, Tns→ 0 as s → , where s is an integer.
135. f 1 0, f 2 1, f 3 2, f 4 4, . . .
134. The series is telescoping:
Sn n
k1
k1
6k
3
3
n
k1
k
2k1
3k
3k
3k1
k1
k
2
3
2k1
3n1
3 n1
3
2n1
lim Sn 3 1 2
In general: f n 2k
n24,
n 14, n odd.
n even
2
(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 n→ x y2 x y2
xy.
4
4
If both odd,
f x y f x y x y2 1 x y2 1
4
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 k24 and
f k 1 f k k
k2
k
2
4
2
k2 2k k 12 1
.
4
4
The argument is similar if k is odd.
260
Chapter 9
Infinite Series
Section 9.3
1.
The Integral Test and p-Series
1
n1
2.
n1
Let f x 1
1
.
x1
Let f x 1
dx lnx 1
x1
1
2
.
3x 5
f is positive, continuous, and decreasing for x ≥ 1.
1
Diverges by Theorem 9.10
3.
2
n1
f is positive, continuous, and decreasing for x ≥ 1.
3n 5
n
4.
ne
n2
n1
n1
Let f x ex.
Let f x xex2.
f is positive, continuous, and decreasing for x ≥ 1.
ex dx ex
1
1
1
Diverges by Theorem 9.10
e
2
2
dx ln3x 5
3x 5
3
f is positive, continuous, and decreasing for x ≥ 3 since
2x
f x < 0 for x ≥ 3.
2e x2
1
e
Converges by Theorem 9.10
xex2 dx 2x 2ex2
3
3
10e32
Converges by Theorem 9.10
5.
n
n1
2
1
1
Let f x 6.
1
.
x2 1
1
x2
1
dx arctan x
1
1
4
1
lnn 1
n1
n1
8.
1
1
ln n
n
n2
lnx 1
.
x1
Let f x f is positive, continuous, and decreasing for x ≥ 2 since
f x 1
dx ln 2x 1
2x 1
Diverges by Theorem 9.10
Let f x 1
.
2x 1
f is positive, continuous, and decreasing for x ≥ 1.
Converges by Theorem 9.10
7.
1
Let f x f is positive, continuous, and decreasing for x ≥ 1.
2n 1
n1
1 lnx 1
< 0 for x ≥ 2.
x 12
lnx 1
lnx 1
2
dx x1
2
Diverges by Theorem 9.10
1
ln x
2 ln x
, f x .
2x32
x
f is positive, continuous, and decreasing for x > e2 7.4.
2
ln x
dx 2xln x 2
x
2
, diverges
Hence, the series diverges by Theorem 9.10.
Section 9.3
9.
1
n n 1
10.
n1
Let f x 1
x x 1
, f x 1 2x
< 0.
x 12
1
1
dx 2 lnx 1
xx 1
1
2
n
3
Let f x 2x32
x
.
x2 3
f x is positive, continuous, and decreasing for x ≥ 2
since
f is positive, continuous, and decreasing for x ≥ 1.
n
n1
The Integral Test and p-Series
, diverges
3 x2
< 0 for x ≥ 2.
x 2 3
fx Hence, the series diverges by Theorem 9.10.
1
x
dx lnx2 3
3
x2
1
Diverges by Theorem 9.10
11.
1
n 1
12.
n1
Let f x n2
1
x 1
1
< 0.
2x 132
, f x 1
1
x 1
dx 2x 1
1
ln n
n3
Let f x f is positive, continuous, and decreasing for x ≥ 1.
ln x
1 3 ln x
, f x .
x3
x4
f is positive, continuous, and decreasing for x > 2.
, diverges
2
2 ln x 1
ln x
dx x2
4x 4
Hence, the series diverges by Theorem 9.10.
2
2 ln 2 1
, converges
16
Hence, the series converges by Theorem 9.10.
13.
ln n
2
n1 n
14.
1
n2
ln x
1 2 ln x
.
Let f x 2 , f x x
x3
Let f x f is positive, continuous, and decreasing for x > e12 1.6.
f is positive, continuous, and decreasing for x ≥ 2.
1
ln x
ln x 1
dx x2
x
1, converges
1
2
Hence, the series converges by Theorem 9.10.
15.
nln n
arctan n
2
n1 n 1
Let f x 1
1
dx 2ln x
xln x
2
, diverges
Hence, the series diverges by Theorem 9.10.
16.
1
n ln n lnln n
n3
arctan x
1 2x arctan x
, f x < 0 for x ≥ 1.
x2 1
x2 12
f is positive, continuous, and decreasing for x ≥ 1.
1
2 ln x 1
, f x 2
.
2x ln x32
xln x
arctan x
arctan x2
dx x2 1
2
1
3 2
, converges
32
Hence, the series converges by Theorem 9.10.
Let f x f x 1
,
x ln x lnln x
ln x 1 lnln x 1
.
x2ln x2lnln x2
f is positive, continuous, and decreasing for x ≥ 3.
3
1
dx lnlnln x
x ln x lnln x
3
, diverges
Hence, the series diverges by Theorem 9.10.
261
262
17.
Chapter 9
n1
Infinite Series
2n
n2 1
Let f x 18.
n1
2x
1
, f x 2 2
< 0 for x > 1.
1
x 12
x2
x2
1
x2
2x
dx lnx 2 1
1
1
, diverges
nk1
k c
n
n1
Let f x 1
20.
1
x
1
dx arctanx 2
1
2
x4
1
, converges
8
ne
k n
n1
x k1
.
xk c
Let f x x k2
ck 1 x k c2
xk
f x < 0
x k1k x
< 0 for x > k.
ex
We use integration by parts.
x k1
xk
.
ex
f is positive, continuous, and decreasing for x > k since
k ck 1.
for x > x
1 3x 4
, f x 4
< 0 for x > 1.
x4 1
x 12
Hence, the series converges by Theorem 9.10.
f is positive, continuous, and decreasing for
k ck 1
x >
since
f x n
1
f is positive, continuous, and decreasing for x > 1.
Hence, the series diverges by Theorem 9.10.
19.
4
Let f x f is positive, continuous, and decreasing for x > 1.
n
1
dx lnx k c
xk c
k
1
1
Diverges by Theorem 9.10
x kex dx x kex
1
k
x k1ex dx
1
1 k kk 1 . . . k!
e
e
e
e
Converges by Theorem 9.10
21. Let f x 1x
, f n an.
x
22. Let f x ex cos x, f n an.
The function f is not positive for x ≥ 1.
The function f is not positive for x ≥ 1.
23. Let f x 2 sin x
, f n an.
x
24. Let f x The function f is not decreasing for x ≥ 1.
25.
1
n
n1
1
1
1
dx 2
x3
2x
1
1
2
Converges by Theorem 9.10
n
1
13
Let f x f is positive, continuous, and decreasing for x ≥ 1.
n
n1
1
Let f x 3.
x
2
The function f is not decreasing for x ≥ 1.
26.
3
sinx x , f n a .
1
.
x13
f is positive, continuous, and decreasing for x ≥ 1.
1
1
13
x
dx 2 x 3
23
1
Diverges by Theorem 9.10
Section 9.3
27.
1
n
1
x
, f x 1
< 0 for x ≥ 1.
2x32
1
1
x
dx 2x
1
1
1
, diverges
1
n n
n1
5
1
n1
Divergent p-series with p 32.
30.
15
1
33.
2
n1
1
n
1
1
1, converges
3
31.
53
34.
32
Convergent p-series with
p 1.04 > 1
1
n
23
n1
1
n
n1
1
12
1
Divergent p-series with p 2 < 1
> 1
1
n
n
n1
5
3
3
Convergent p-series with p 2 > 1
36.
1.04
n1
n
n1
Convergent p-series with p 2 > 1
35.
x
x
dx Convergent p-series with p < 1
n
2
Hence, the series converges by Theorem 9.10.
n1
1
5
1
2
, f x 3 < 0 for x ≥ 1.
x2
x
f is positive, continuous, and decreasing for x ≥ 1.
Hence, the series diverges by Theorem 9.10.
29.
2
Let f x f is positive, continuous, and decreasing for x ≥ 1.
1
n1
Let f x 2
Divergent p-series with p 3 < 1
37.
2
n
34
n1
S1 2
Convergent p-series with
p > 1
S2 3.1892
S3 4.0666
S4 4.7740
Matches (c), diverges
38.
2
n
39.
2
n
2
n
n1
40.
2
2
n
n1
25
n1
n1
S1 2
S1 2
S1 2
S2 3
S2 2.6732
S2 3.5157
S3 3.6667
S3 3.0293
S3 4.8045
S4 4.1667
S4 3.2560
Matches (a), diverges
Matches (f), diverges
Matches (b), converges
Note: The partial sums for 39 and
41 are very similar because
2 32.
41.
2
2
nn n
n1
n1
263
n
28.
n1
The Integral Test and p-Series
32
42.
2
n
n1
2
S1 2
S1 2
S2 2.7071
S2 2.5
S3 3.0920
S3 2.7222
S4 3.3420
Matches (e), converges
Matches (d), converges
Note: The partial sums for 39 and 41 are very similar
because 2 32.
264
Chapter 9
43. (a)
Infinite Series
8
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.
0
11
0
(b)
5
n
5
10
20
50
100
Sn
1.4636
1.5498
1.5962
1.6251
1.635
The partial sums approach the sum 26 1.6449
slower than the series in part (a).
N
44.
1
1
1
1
1
n1234. . .N
0
11
0
> M
n1
(a)
M
2
4
6
8
N
4
31
227
1674
(b) No. Since the terms are decreasing (approaching zero),
more and more terms are required to increase the partial
sum by 2.
45. Let f be positive, continuous, and decreasing for x ≥ 1
and an f n. Then,
a
n
46. A series of the form
n1
n1
1
is a p-series, p > 0.
np
The p-series converges if p > 1 and diverges if
0 < p ≤ 1.
and
f x dx
1
either both converge or both diverge (Theorem 9.10).
See Example 1, page 618.
48. No. Theorem 9.9 says that if the series converges, then
the terms an tend to zero. Some of the series in Exercises
37–42 converge because the terms tend to 0 very rapidly.
47. Your friend is not correct. The series
1
1
1
. . .
n
10,000
10,001
n10,000
is the harmonic series, starting with the 10,000th term,
and hence diverges.
49. The area under the rectangles is greater than the area under the
graph of y 1x, x ≥ 1.
Since
1
1
x
dx
0.5
a
n
≥
n1
f x dx ≥
a
1
x
1
dx 2x
7
6
50.
1.0
1
>
n
n1
y
diverges, the series
1
n diverges.
n
x
n1
7
n2
1
1
51.
1
3
4
5
n
1
nln n
n2
2
p
If p 1, then the series diverges by the Integral Test.
If p 1,
y
1
2
x
1
2
3
4
5
6
7
1
dx xln x p
2
ln xp
1
ln xp1
dx x
p 1
Converges for p 1 < 0 or p > 1
2
.
Section 9.3
52.
The Integral Test and p-Series
265
ln n
p
n2 n
If p 1, then the series diverges by the Integral Test. If p 1,
2
ln x
dx xp
xp ln x dx 2
xp1
p 1
2
1 p 1 ln x
. (Use integration by parts.)
2
Converges for p 1 < 0 or p > 1
53.
n
1 n 2 p
n1
If p 1,
n
1n
f x diverges (see Example 1). Let
2
n1
x
, p1
1 x2 p
f x 1 2p 1x2
.
1 x2 p1
For a fixed p > 0, p 1, f x is eventually negative. f is positive, continuous, and eventually decreasing.
1
x
1
dx 1 x2 p
x2 1 p12 2p
1
For p > 1, this integral converges. For 0 < p < 1, it diverges.
54.
n1 n 2 p
55.
n1
1
nln n
n2
Since p > 0, the series diverges for all values of p.
Hence,
p
converges for p > 1.
1
n ln n, diverges.
n2
56.
1
nln n
n2
Hence,
p
1
nln n
3
1
nln n
n2
Hence,
57.
p
2
1
nln n
n2
23 ,
diverges.
converges for p > 1.
1
1
2
n2
1
1 , diverges
2 n2 n lnn
Hence,
p
converges for p > 1.
1
nln n , converges.
n2
2
59. Since f is positive, continuous, and decreasing for x ≥ 1
and an f n, we have,
1
n lnn 2n ln n
n2
nln n
n2
n2
58.
converges for p > 1.
RN S SN a
n
n1
N
a
n
n1
an ≤ aN1 nN1
≤
Thus, 0 ≤ RN ≤
N
f x dx.
f x dx
N1
N
an > 0.
nN1
Also,
RN S SN f x dx.
266
Chapter 9
Infinite Series
61. S6 1 60. From Exercise 59, we have:
0 ≤ S SN ≤
f x dx
R6 ≤
N
SN ≤ S ≤ SN N
n
≤ S ≤
n1
N
a
n
f x dx
n1
62. S4 1 R4 ≤
4
1
n
5
≤ 1.0811 0.0015 1.0826
4
f x dx
4
0.0010
1
dx arctan x
x 1
n
0.9818 ≤
n1
2
10
arctan 10 0.0997
2
1
≤ 0.9818 0.0997 1.0815
1
1
1
1
1
. . .
1.9821
2ln 23 3ln 33 4ln 43
11ln 113
1
1
dx x 1lnx 1
3
2lnx 1
2
10
1
n 1lnn 1
1.9821 ≤
3
n1
R4 ≤
0.0015
≤ 1.0363 0.0010 1.0373
2
10
65. S4 1
n1
6
1
1 1
1
1
1
1
1
1
1
0.9818
2 5 10 17 26 37 50 65 82 101
R10 ≤
n
1.0811 ≤
N
64. S10 1
1
dx 4
x5
4x
n1
R10 ≤
1
1
dx 3
x4
3x
1
1
1
1.0363
25 35 45
1.0363 ≤
63. S10 6
N
a
1
1
1
1
1
1.0811
24 34 44 54 64
10
4
nen
0.4049 ≤
2
1
2
2
xex d x ex
2
4
1
0.0870
2ln 113
≤ 1.9821 0.0870 2.0691
1
2
3
4
4 9 16 0.4049
e
e
e
e
66. S4 e 16
5.6 108
2
≤ 0.4049 5.6 108
R4 ≤
1
1
1
1
2 3 4 0.5713
e
e
e
e
4
0.5713 ≤
N
1
< 0.003
N3
N 3 > 333.33
N > 6.93
N ≥ 7
e
n
0.0183
4
≤ 0.5713 0.0183 0.5896
n0
n1
67. 0 ≤ RN ≤
ex dx ex
1
1
dx 3
x4
3x
N
1
< 0.001
3N 3
68. 0 ≤ RN ≤
N
1
32
x
N12 < 0.0005
N > 2000
N ≥ 4,000,000
dx 2
12
x
N
2
N
< 0.001
Section 9.3
69. RN ≤
N
1
e5x dx e5x
5
N
e5N
< 0.001
5
RN ≤
70.
ex2 dx 2ex2
N
1
< 0.005
e5N
2
< 0.001
e N2
e5N > 200
e N2 > 2000
5N > ln 200
N
267
2
< 0.001
e N2
N
> ln 2000
2
ln 200
5
N >
The Integral Test and p-Series
N > 2 ln 2000 15.2
N ≥ 16
N > 1.0597
N ≥ 2
71. RN ≤
N
1
dx arctan x
x2 1
72. Rn ≤
N
N
arctan N < 0.001
2
2
1
x
arctan
dx 2
x2 5
5
5
N
< 0.001
2 N
arctan
2
5
5
arctan N < 1.5698
N
arctan
< 0.001118
2
5
arctan N > 1.5698
N > tan 1.5698
1.56968 < arctan
N ≥ 1004
N
5
N5
> tan 1.56968
N ≥ 2004
73. (a)
1
n
n2
1.1 .
f x 2
6
(b)
6
1
n ln n is a divergent series. Use the Integral Test.
n2
1
is positive, continuous, and decreasing for x ≥ 2.
x ln x
1
dx ln ln x
x ln x
1
n
n2
This is a convergent p-series with p 1.1 > 1.
1.1
2
1
1
1
1
1
0.4665 0.2987 0.2176 0.1703 0.1393
21.1 31.1 41.1 51.1 61.1
1
1
1
1
1
1
n ln n 2 ln 2 3 ln 3 4 ln 4 5 ln 5 6 ln 6 0.7213 0.3034 0.1803 0.1243 0.0930
n2
For n ≥ 4, the terms of the convergent series seem to be larger than those of the divergent series!
1
1
<
n1.1
n ln n
(c)
n ln n < n1.1
ln n < n0.1
This inequality holds when n ≥ 3.5 1015. Or, n > e40. Then ln e40 40 < e400.1 e4 55.
74. (a)
10
xp1
1
dx p
x
p 1
10
1
, p > 1
p 110 p1
(c) The horizontal asymptote is y 0. As n increases, the
error decreases.
(b) f x 1
xp
R10 p n11
1
np
≤ Area under the graph of f over the interval 10, 268
Chapter 9
Infinite Series
75. (a) Let f x 1x. f is positive, continuous, and decreasing on 1, .
Sn 1 ≤
n
1
y
1
dx
x
1
Sn 1 ≤ ln n
1
2
Hence, Sn ≤ 1 ln n. Similarly,
Sn ≥
n1
1
1
dx lnn 1.
x
1
3 ...
n −1
2
x
n
n+1
Thus, lnn 1 ≤ Sn ≤ 1 ln n.
(b) Since lnn 1 ≤ Sn ≤ 1 ln n, we have lnn 1 ln n ≤ Sn ln n ≤ 1. Also, since ln x is an increasing function,
lnn 1 ln n > 0 for n ≥ 1. Thus, 0 ≤ Sn ln n ≤ 1 and the sequence an is bounded.
(c) an an1 Sn ln n
Sn1 lnn 1
n1
n
1
1
dx ≥ 0
x
n1
Thus, an ≥ an1 and the sequence is decreasing.
(d) Since the sequence is bounded and monotonic, it converges to a limit, .
(e) a100 S100 ln 100 0.5822 (Actually 0.577216.)
76.
ln1 n ln
1
2
n2
n2
n2 1
n 1(n 1
ln
lnn 1 lnn 1 2 ln n
2
2
n
n
n2
n2
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 . . . ln 2
77.
x
ln n
n2
1
(a) x 1:
n2
n2
1, diverges
(c) Let x be given, x > 0. Put x ep ⇔ ln x p.
n2
1
(b) x :
e
ln n
1
e
ln n
eln n n2
x
1
, diverges
n2 n
ln n
n2
e
p ln n
n2
n
p
n2
1
n
n2
p
1
This series converges for p > 1 ⇒ x < .
e
78. x n
x
n1
1
n
n1
x
Converges for x > 1 by Theorem 9.11
79.
1
2n 1
80.
n1
Let f x n2
1
.
2x 1
1
1
dx ln 2x 1
2x 1
Diverges by Theorem 9.10
1
2
Let f x f is positive, continuous, and decreasing for x ≥ 1.
nn
1
1
1
.
xx 2 1
f is positive, continuous, and decreasing for x ≥ 2.
2
1
xx 2 1
Converges by Theorem 9.10
dx arcsec x
2
2
3
Section 9.3
81.
nn n
1
p-series with p 5
4
1
4
n1
n1
82. 3
54
2
3
0.95
p-series with p 0.95
Diverges by Theorem 9.11
n
84.
1.075
n
n0
n0
Geometric series with r 3
Geometric series with r 1.075
Converges by Theorem 9.6
Diverges by Theorem 9.6
2
85.
n
n
1
86.
2
n1
n
n1
n
lim
n→ n2 1
1
lim
1 1n2
n→ 1
2
1
1
1
2 3
n3
n
n1
n1 n
Since these are both convergent p-series, the difference
is convergent.
10
Diverges by Theorem 9.9
87.
1 n 1
n
88.
1
n
lim 1 lnn
n2
n1
n→ n
lim lnn e0
n→ Diverges by Theorem 9.9
Fails nth-Term Test
Diverges by Theorem 9.9
89.
1
nln n
3
n2
Let f x 1
.
xln x3
f is positive, continuous and decreasing for x ≥ 2.
2
1
dx xln x3
ln x3
2
1
ln x2
dx x
2
2
1
2ln x2
2
1
2ln 22
Converges by Theorem 9.10. See Exercise 51.
90.
n2
ln n
n3
Let f x ln x
.
x3
f is positive, continuous, and decreasing for x ≥ 2 since fx 2
269
1
n
n1
Converges by Theorem 9.11
83.
The Integral Test and p-Series
ln x
ln x
dx 2
x3
2x
2
1
2
2
1
ln 2
1
2
dx x3
8
4x
Converges by Theorem 9.10. See Exercise 30.
2
1 3 ln x
< 0 for x ≥ 2.
x4
1
ln 2
(Use integration by parts.)
8
16
270
Chapter 9
Infinite Series
Section 9.4
1. (a)
6
n
n1
32
6
6
. . . ; S1 6
1 232
an
6
an = 3/2
n
6
n
n1
Comparisons of Series
6
6
6
3
. . . ; S1 3 4 232 3
2
5
32
an =
4
3
6
6
6
6
. . . ; S1 4.9
2 0.5
2
n
n
1
1.5
4.5
1.5
n1
(b) The first series is a p-series. It converges p an = 3/26
n +3
2
1
n
> 1.
3
2
6
n n 2 + 0.5
(c) The magnitude of the terms of the other two series are less than the corresponding
terms at the convergent p-series. Hence, the other two series converge.
6
4
2
Sn
n
Σ
k=1
12
8
10
6
k 3/2
n
Σ
10
(d) The smaller the magnitude of the terms, the smaller the magnitude of the terms
of the sequence of partial sums.
k=1
6
k k 2+ 0.5
8
6
4
n
Σ
2
6
k 3/2 + 3
k=1
n
2. (a)
2
2
n 2 2 . . . S
1
2
an
an =
n1
6
4
2
8
10
2
n − 0.5
4
2
2
2
. . . S1 4
0.5
2 0.5
n1 n 0.5
3
2
4
4
4
. . . S1 3.3
1.5
2.5
n1 n 0.5
(b) The first series is a p-series. It diverges p 1
2
4
n + 0.5
an =
an = 2
n
1
< 1.
n
2
(c) The magnitude of the terms of the other two series are greater than the corresponding
terms of the divergent p-series. Hence, the other two series diverge.
4
6
Sn
20
(d) The larger the magnitude of the terms, the larger the magnitude of the terms of the
sequence of partial sums.
Σ
16
2
n − 0.5
8
10
Σ
4
n + 0.5
12
8
4
Σ
2
n
8
10
n
2
3. 0 <
1
1
< 2
n2 1
n
Therefore,
n
n1
2
1
1
converges by comparison with the
convergent p-series
1
n.
n1
2
4.
1
1
<
3n2 2
3n2
Therefore,
3n
n1
2
1
2
5.
4
6
1
1
> > 0 for n ≥ 2
n1
n
Therefore,
1
n1
n2
converges by comparison with the
convergent p-series
diverges by comparison with the
divergent p-series
1 1
.
3 n1 n2
n.
n2
1
Section 9.4
6.
1
n 1
>
1
n
for n ≥ 2
7. 0 <
Therefore,
1
1
< n
3n 1
3
8.
Therefore,
1
n 1
n
n0
diverges by comparison with the
divergent p-series
converges by comparison with the
convergent geometric series
n2
n
1
ln n
1
>
> 0.
n1
n1
10.
1
n3 1
ln n
n
n1
<
1
n32
1
1
diverges by comparison with the
divergent series
n
n1
n1
1
n!
3
n1
1
1
1
>
> 0.
n2
n!
Therefore,
converges by comparison with the
convergent p-series
3 n
.
4
11. For n > 3,
n1
n 1.
3n
4 5
n
n0
Therefore,
Therefore,
n
converges by comparison with the
convergent geometric series
n0
9. For n ≥ 3,
n0
3 .
1
3n
3
<
5
4
n2
n.
4n
271
Therefore,
1
1
3
Comparisons of Series
n0
converges by comparison with the
convergent p-series
1
1
n.
32 .
2
n1
1
n 1 diverges by the
Note:
n1
Integral Test.
12.
1
1
3 n
4
>
1
4 n
4
Therefore,
1
3
n1 4 n 1
diverges by comparison with the
divergent p-series
1 1
.
4 n
4 n1 13. 0 <
1
1
≤ n
e n2
e
Therefore,
n1
1
n.
n1
4n
1
n1
converges by comparison with the
convergent geometric series
diverges by comparison with the
divergent geometric series
e .
1
3 .
n
n0
n
1
diverges by a limit comparison with the divergent p-series
n
n0
Therefore,
2
4n
4n
> n
1
3
3
2
nn2 1
n2
lim 2
1
n→ n→ n 1
1n
3n
Therefore,
1
en
15. lim
n
14.
4
n
n1
23n 5
2 3n
lim n
2
n
n→ n→ 3 5
13
16. lim
Therefore,
3
n1
n
2
5
converges by a limit comparison with the convergent
geometric series
3 .
n1
1
n
272
Chapter 9
Infinite Series
1n2 1
n
lim
1
n→ n2 1
1n
17. lim
n→ Therefore,
n
1
1
diverges by a limit comparison with the divergent p-series
n→ Therefore,
2
n0
3n2 4
3n
3
lim
n→ n2 4
1n
18. lim
1
n.
n
3
4
2
n3
diverges by a limit comparison with the divergent
harmonic series
1
n.
n1
n3
2n2 1
2
2n 1
2n5 n3
lim
3
5
n→ 3n 2n 1
1n
3
3n5
19. lim
n→ Therefore,
Therefore,
3n
2n2 1
2n 1
5
n1
converges by a limit comparison with the convergent
p-series
1
.
3
n1 n
n
2
n1
5n 3
2n 5
diverges by a limit comparison with the divergent p-series
1
n.
n1
n3
n2 3n
nn 2
lim 2
1
21. lim
n→ n→
1n
n 2n
1
n3
nn2 1
22. lim
lim 3
1
3
n→ n→
1n
n n
Therefore,
Therefore,
n3
5n 3
n2 2n 5
5n2 3n
20. lim
lim 2
5
n→ n→ n 2n 5
1n
nn 2
nn
2
1
1
n1
n1
diverges by a limit comparison with the divergent p-series
converges by a limit comparison with the convergent
p-series
1
n.
n1
23. lim
n→ 1nn2 1
n2
lim
1
n→ nn2 1
1n2
2
n1
1
converges by a limit comparison with the convergent
p-series
1
n.
n1
2
n→ nn 12n1
n
lim
1
n→ n 1
12n1
1
nn
24. lim
3
Therefore,
Therefore,
1
n.
n1
n
n 12
n1
n1
converges by a limit comparison with the convergent
geometric series
2
n1
1
n1
.
Section 9.4
25. lim
n→ nk1nk 1
nk
lim k
1
n→ n 1
1n
n→ Therefore,
nk1
n
2
n1
diverges by a limit comparison with the divergent p-series
5
n n
1
k
n1
1
1
.
n
n1
n1
sin1n
1n2 cos1n
lim
n→ n→ 1n
1n2
tan1n
1n2 sec21n
lim
n→ n→ 1n
1n2
27. lim
lim cos
n→ 28. lim
1n 1
lim sec2
n→ Therefore,
1n 1
Therefore,
sinn
tann
1
1
n1
n1
diverges by a limit comparison with the divergent p-series
diverges by a limit comparison with the divergent p-series
1
n.
n1
n
n1
1
n.
n1
29.
4
diverges by a limit comparison with the divergent
harmonic series
n.
n
1
n
30.
n1
Diverges
1
p-series with p 2
5 5 1
n
3
31.
n
1
2
n0
n1
Converges
Converges
1
Geometric series with r 5
Direct comparison with
3
n1
32.
3n
2
n4
1
2n 15
33.
n
Diverges; nth-Term Test
Limit comparison with
1
n
n1
2n 3
n1
Converges
34.
lim
2
n→ n
1
0
2n 3 2
n 1 n 2 2 3 3 4 4 5 . . . 2
1
1
1
1
1
1
1
1
1
n1
Converges; telescoping series
35.
273
5n n2 4
5n
5
lim
n→ n n2 4
2
1n
26. lim
Therefore,
Comparisons of Series
n
n1
2
n
12
Converges; Integral Test
36.
3
nn 3
n1
Converges; telescoping series
n n 3
n1
1
1
1
n
274
Chapter 9
Infinite Series
an
lim na . By given conditions lim nan
n→ 1n n→ n
is finite and nonzero. Therefore,
37. lim
n→ 38. If j < k 1, then k j > 1. The p-series with
p k j converges and since
lim
n→ an
Pn
PnQn
L > 0, the series
1nkj
n1 Qn
n1
converges by the Limit Comparison Test. Similarly, if
j ≥ k 1, then k j ≤ 1 which implies that
diverges by a limit comparison with the p-series
1
n.
Pn
Qn
n1
n1
diverges by the Limit Comparison Test.
39.
3
4
5
1 2
n
. . .
,
2 1
2 5 10 17 26
n
n1
40.
which diverges since the degree of the numerator is only
one less than the degree of the denominator.
41.
n
n1
3
1
1
42.
5n n 3 lim
4
Therefore,
n→ 5n
n1
45.
3
n4
1
0
3 5
5n4
44. lim
n→ n3
diverges.
3
n
1
lim
lim n 0
ln n n→ 1n n→
Therefore,
4
46.
1
1
1
1
. . .
200 210 220
200
10n
n0
48.
1
1
1
1
1
. . .
3
201 208 227 264
n1 200 n
converges
converges
49. Some series diverge or converge very slowly. You
cannot decide convergence or divergence of a series
by comparing the first few terms.
50. See Theorem 9.12, page 624. One example is
n
2
n1
1
n
n1
51. See Theorem 9.13, page 626. One example is
1
n 1 diverges because lim
1
diverges ( p-series).
n2 n
1
diverges
1
1
1
1
1
201 204 209 216 n1 200 n2
n2
ln n diverges.
n2
1
1
1
1
. . .
200 400 600
200n
n1
n2
1
diverges since the degree of the numerator is only one less
than the degree of the denominator.
diverges, (harmonic)
47.
n
n1
3
n→ which converges since the degree of the numerator is two
less than the degree of the denominator.
converges since the degree of the numerator is three less
than the degree of the denominator.
43. lim n
1
1
1
1
1 1
. . .
,
2 1
3 8 15 24 35
n
n2
n→
1n 1
1 and
1n
2
1
1
1
converges because 2
<
and
1
n 1 n2
converges ( p-series).
52. This is not correct. The beginning terms do not affect the
convergence or divergence of a series. In fact,
1
1
1
. . .
diverges (harmonic)
1000 1001
n
n1000
and 1 1
1 1 . . .
2 converges ( p-series).
4 9
n1 n
Section 9.4
53.
1.0
Comparisons of Series
275
Terms of
∞
Σ an
0.8
n=1
0.6
Terms of
∞
2
Σ an
0.4
n=1
0.2
n
4
8
12
16
20
For 0 < an < 1, 0 < an2 < an < 1. Hence, the lower terms are those of an2.
1
2n 1
54. (a)
2
n1
1
4n 1
4n
n1
2
(b)
converges since the degree of the numerator is
two less than the degree of the denominator.
(See Exercise 38.)
1
2n 1
(c)
2
n3
55. False. Let an (d)
2
S2 0.1226
8
n
5
10
20
50
100
Sn
1.1839
1.02087
1.2212
1.2287
1.2312
1
2
S9 0.0277
2
8
n10 2n 1
1
1
1
1
and bn 2. 0 < an ≤ bn and both
and
converge.
3
3
2
n
n
n1 n
n1 n
56. True
57. True
58. False. Let an 1n, bn 1n, cn 1n2. Then,
an ≤ bn cn, but
c
n
59. True
converges.
n1
60. False.
a
n
61. Since
could converge or diverge.
For example, let
b
n
n1
1
n, which diverges.
n1
ab
n
series
a
n
converges, then
n1
a
n an
1
2
n1
converge, and hence so does
converges by comparison to the convergent
n
a
n.
i1
a
2
63.
n
n1
1
n
2
1
n
and
3
both converge, and hence so does
n n n .
converges by Exercise 61.
n
n→ n1
1
1
1
0 < 2 <
and
2 converges.
n
n
n1 n
64.
converges, lim bn 0. There exists N such
that bn < 1 for n > N. Thus, anbn < an for n > N and
1
1
1
<
and
diverges, but
n n
n1 n
62. Since
n
n1
n1
0 <
b
n 1
2
2
1
n .
4
1
1
1
2
3
5
65. Suppose lim
n→ an
0 and bn converges.
bn
From the definition of limit of a sequence, there exists
M > 0 such that
an
0 < 1
bn
whenever n > M. Hence, an < bn for n > M. From the
Comparison Test, an converges.
276
Chapter 9
66. Suppose lim
n→ Infinite Series
an
and bn diverges. From the definition of limit of a sequence, there exists M > 0 such that
bn an
> 1
bn
for n > M. Thus, an > bn for n > M. By the Comparison Test, an diverges.
a
67. (a) Let
n
lim
n→ 1
n 1 , and b
n
3
1
n 1
a
lim
2
a
(b) Let
n
n
ln n
, and
n
lim
n→ converges.
3
n1
n→ 1
n , converges.
an
1n 13
n2
lim
lim
0
n→ n 13
bn n→
1n2
By Exercise 65,
68. (a) Let
n1
b
n
1
n, diverges.
(b) Let
sinan
1 and
n→ an
a
n
1
n
n
1
ln n, and b
n
By Exercise 66,
70.
1
, converges.
n
converges.
1
n, diverges.
an
n
lim
bn n→ ln n lim
n→ ln n
diverges.
n
an
1n n
1
lim
lim
0
n→ n
bn n→ 1 n
n1
69. Since lim an 0, the terms of sinan are positive for
n→ sufficiently large n. Since
lim
n
n
By Exercise 65,
an
ln nn
lim ln n lim
n→ bn n→ 1n
By Exercise 66,
1
n , and b
1
ln n diverges.
1
1
1 2 . . . n nn 12
n1
n1
an
2
nn 1
n1
Since 1n2 converges, and
converges, so does sinan.
lim
n→ 2nn 1
2n2
lim
2,
n→ nn 1
1n2
1
1 2 . . . n converges.
72. Consider two cases:
71. The series diverges. For n > 1,
n < 2n
n1n
1
n1n
1
nn1n
Since
If an ≥
< 2
>
annn1 1
2
If an ≤
1
>
2n
1
diverges, so does
2n
n
1
n1n .
1n1
2 1 nn1
1
1
, then an1n1 ≥ n1
2n1
2
1
, and
2
an
≤ 2an.
an1n1
1
n1 ,
2
then annn1 ≤
combining, annn1 ≤ 2an 1
, and
2n
1
.
2n
1
annn1 by
n converges, so does
2
n1
n1
the Comparison Test.
Since
n1
2a
n
Section 9.5
Section 9.5
1.
4.
Alternating Series
6
n
2.
2
n1
Alternating Series
1n16
n2
n1
3.
3
n!
n1
S1 6
S1 6
S1 3
S2 7.5
S2 4.4
S2 4.5
S3 8.1667
S3 5.1667
S3 5.0
Matches (d).
Matches (f ).
Matches (a).
1n13
n!
n1
5.
10
n2
n1
6.
n
1n10
n2n
n1
S1 3
S1 5
S1 5
S2 1.5
S2 6.25
S2 3.75
S3 2.0
Matches (e).
Matches (c).
Matches (b).
7.
1n1 0.7854
4
n1 2n 1
(a)
(b)
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
0
11
0.6
(c) The points alternate sides of the horizontal line 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.
8.
1n1
1
n 1! e 0.3679
n1
(a)
(b)
n
1
2
3
4
5
Sn
1
0
0.5
0.3333
0.375
6
0.3667
7
8
9
10
0.3681
0.3679
0.3679
0.3679
2
0
11
0
(c) The points alternate sides of the horizontal line 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.
277
278
9.
Chapter 9
Infinite Series
1n1 2
0.8225
n2
12
n1
(a)
(b)
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 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.
1n1
2n 1! sin1 0.8415
n1
(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 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.
11.
1n1
n
n1
an1 lim
n→ 12.
1
1
< an
n1
n
1n1 n
n1 2n 1
lim
n→ 1
0
n
n
1
2n 1 2
Diverges by the nth-Term Test
Converges by Theorem 9.14
13.
1n1
n1 2n 1
an1 lim
n→ 1
1
<
an
2n 1 1
2n 1
1
0
2n 1
Converges by Theorem 9.14
14.
1n
lnn 1
n1
an1 lim
n→ 1
1
<
an
lnn 2 lnn 1
1
0
lnn 1
Converges by Theorem 9.14
Section 9.5
15.
1n n2
2
n1 n 1
16.
n2
1. Thus, lim an 0.
n→ n 1
n→ lim
1n1n
2
n1 n 1
an1 2
Diverges by the nth-Term Test
Alternating Series
lim
n→ n
n1
< 2
an
n 12 1
n 1
n
0
n2 1
Converges by Theorem 9.14
17.
1n
n1 n
an1 lim
n→ 18.
1
1
<
an
n 1
n
1
n
1n1n2
2
n1 n 5
lim
n→ n2
n2
1
5
Diverges by nth-Term Test
0
Converges by Theorem 9.14
19.
1n1n 1
lnn 1
n1
lim
n→ 20.
n1
1
lim n 1 lim
lnn 1 n→ 1n 1 n→
1n1 lnn 1
n1
n1
an1 Diverges by the nth-Term Test
lim
n→ lnn 1 1
lnn 1
<
for n ≥ 2
n 1 1
n1
1n 1
lnn 1
lim
0
n→ n1
1
Converges by Theorem 9.14
21.
sin
n1
2n 1
1n1
2
n1
22.
n1
n1
cos n 1
n
1
n1
Diverges by the nth-Term Test
23.
n sin
1n1
2n 1
2
n
n1
Converges; (See Exercise 11.)
24.
1n
n
n1
1
n cos n n1
Diverges by the nth-Term Test
25.
Converges; (See Exercise 11.)
1n
n!
n0
an1 lim
n→ 1
1
<
an
n 1!
n!
1
0
n!
Converges by Theorem 9.14
26.
1n
2n 1!
n0
an1 lim
n→ 1
1
<
an
2n 3! 2n 1!
27.
1n1n
n2
n1
an1 1
0
2n 1!
Converges by Theorem 9.14
<
lim
n→ n 1
n 1 2
n
for n ≥ 2
n2
n
0
n2
Converges by Theorem 9.14
28.
1n1n
3 n
n1
lim
n→ n12
lim n16 n13 n→
Diverges by the nth-Term Test
279
280
29.
Chapter 9
Infinite Series
1n1n!
1 3 5 . . . 2n 1
n1
an1 1
35.
lim an lim
n→ n→ lim
n→ n 1!
. . 2n 12n 1 1
35.
1
123. . .n
3 5 . . . 2n 1
n→ n!
. . 2n 1
n1
n1
2n 1 an 2n 1 < an
n!
. . 2n 1
1
lim 2
35.
33 45 57 . . . 2n n 3
2n 1 1 0
Converges by Theorem 9.14
30.
1
n1
n1
an1 1
1
13
14
35.
47.
5.
7.
.
.
. 2n 1
. 3n 2
. . 2n 12n 1
2n 1
. . 3n 23n 1 an 3n 1 < an
1
1
0
54 77 109 . . . 2n
3n 5
3n 2
lim an lim 3
n→ n→ Converges by Theorem 9.14
31.
1n12en
1n12
n
n
e2n 1
n1 e e
n1
Let f x 2ex
. Then
e2x 1
lim
n→ e2n 1
lim
n→ 2en
2e2n
lim
n→ 1
0.
en
The series converges by Theorem 9.14.
2ex
. Then
e2x 1
2e2x 1 e2x
< 0 for x > 0.
e2x 12
f x Thus, f x is decreasing. Therefore, an1 < an , and
2en
1n12en
21n1
n
n
e2n 1
n1 e e
n1
Let f x 2ex e2x 1
< 0.
e2x 12
f x 32.
Thus, f x is decreasing for x > 0 which implies
an1 < an.
lim
n→ 2en
2en
1
lim 2n lim n 0
n→ e
1 n→ 2e
e2n
The series converges by Theorem 9.14.
33. S6 31n1
2.4325
n2
n1
6
2.4325 0.0612 ≤ S ≤ 2.4325 0.0612
3
R6 S S6 ≤ a7 49 0.0612
34. S6 6
41n1
lnn 1 2.7067
n1
4
R6 S S6 ≤ a7 ln 8 1.9236
0.7831 ≤ S ≤ 4.6303
2.3713 ≤ S ≤ 2.4937
35. S6 21n
0.7333
n!
n0
5
0.7333 0.002778 ≤ S ≤ 0.7333 0.002778
2
R6 S S6 ≤ a7 6! 0.002778
0.7305 ≤ S ≤ 0.7361
36. S6 1n1n
0.1875
2n
n1
6
7
R6 S S6 ≤ a7 27 0.05469
0.1328 ≤ S ≤ 0.2422
Section 9.5
37.
1n
n!
n0
38.
(a) By Theorem 9.15,
RN ≤ aN1 1n
n
n0 2 n!
(a) By Theorem 9.15,
1
< 0.001.
N 1!
Rn
This inequality is valid when N 6.
(b) We may approximate the series by
1
1 1
1
1
1
11 n!
2 6 24 120 720
n0
1n
1 1
1
1
1 0.607.
n
2 8 48 348
n0 2 n!
n
1
< 0.001.
2N1N 1!
≤ aN1 This inequality is valid when N 4.
(b) We may approximate the series by
6
Alternating Series
4
0.368.
(5 terms. Note that the sum begins with n 0.)
(7 terms. Note that the sum begins with n 0.)
39.
1n
n0
1n
n0 2n!
(a) By Theorem 9.15,
(a) By Theorem 9.15,
2n 1!
RN 40.
≤ aN1 1
< 0.001.
2N 1 1!
RN This inequality is valid when N 2.
(b) We may approximate the series by
2n 1! 1 6 120 0.842.
n0
1n
1
1
1
1 0.540.
2n
!
2
24
720
n0
(3 terms. Note that the sum begins with n 0.)
(4 terms. Note that the sum begins with n 0.)
2
41.
1
< 0.001.
2N 2!
This inequality is valid when N 3.
(b) We may approximate the series by
1n
≤ aN1 1
3
1
1n1
n
n1
42.
(a) By Theorem 9.15,
RN ≤ aN1 1n1
4n n
n1
(a) By Theorem 9.15,
1
< 0.001.
N1
RN This inequality is valid when N 1000.
≤ aN1 1
< 0.001.
4N1N 1
This inequality is valid when N 3.
(b) We may approximate the series by
(b) We may approximate the series by
1n1
1 1 1
1
1 . . .
n
2
3
4
1000
n1
1n1 1
1
1
0.224.
nn
4
4
32
192
n1
1000
3
0.693.
(3 terms)
(1000 terms)
43.
1n1
n3
n1
44.
By Theorem 9.15,
By Theorem 9.15,
1
RN ≤ aN1 N 13 < 0.001
⇒ N 13 > 1000 ⇒ N 1 > 10.
Use 10 terms.
1n1
n2
n1
1
RN ≤ aN1 N 12 < 0.001
⇒ N 12 > 1000 ⇒ N 31.
Use 31 terms.
281
282
45.
Chapter 9
Infinite Series
1n1
3
n1 2n 1
46.
By Theorem 9.15,
RN ≤ aN1 1n1
n4
n1
By Theorem 9.15, RN ≤ aN1 1
< 0.001.
2N 13 1
1
< 0.001.
N 14
This inequality is valid when N 5.
This inequality is valid when N 7. Use 7 terms.
47.
1n1
n 1
n1
1
n 1
2
n1
48.
2
The given series converges by the Alternating Series Test,
but does not converge absolutely since the series
converges by comparison to the p-series
2
n1
diverges by the Integral Test. Therefore, the series
converges conditionally.
Therefore, the given series converges absolutely.
49.
1n1
n
n1
50.
1n1
nn
n1
The given series converges by the Alternating Series Test,
but does not converge absolutely since
1
n1
1
n.
n1
1n1
n1 n 1
1
1
nn n
n1
1
n1
32
which is a convergent p-series.
Therefore, the given series converges absolutely.
n
n1
is a divergent p-series. Therefore, the series converges
conditionally.
51.
1n1 n2
2
n1 n 1
52.
n→ Therefore, the series diverges by the nth-Term Test.
Therefore, the series diverges by the nth-Term Test.
lim
1n
n2 ln n
54.
1
diverges by comparison to the harmonic series
ln
n
n 2
1
. Therefore, the series converges conditionally.
n1 n
1n n
3
n2 n 1
n
converges by a limit comparison to
1
n2
1
.
the convergent p-series
2 Therefore, the given series
n2 n
converges absolutely.
3
converges by a comparison to
e . Therefore, the
1
n
n0
given series converges absolutely.
56.
n
2
the convergent geometric series
1
en
n0
1n
n2
n0 e
The given series converges by the Alternating Series Test
but does not converge absolutely since the series
55.
2n 3
2
n 10
n2
1
n→ n 12
lim
53.
1n12n 3
n 10
n1
1n1
n1.5
n1
1
n
n1
1.5
is a convergent p-series.
Therefore, the given series converges absolutely.
Section 9.5
57.
1n
2n 1!
58.
n0
Alternating Series
283
1n
n 4
n0
1
n0
The given series converges by the Alternating Series Test,
but
is convergent by comparison to the convergent geometric
series
n0 n
2n 1!
2
1
1
4
n
diverges by a limit comparison to the divergent p-series
n0
1
n.
since
n1
1
1
< n for n > 0.
2n 1!
2
Therefore, the given series converges conditionally.
Therefore, the given series converges absolutely.
59.
1n
cos n
n0 n 1
n0 n 1
60.
cos n n1
n0
n1
arctan n
n1
The given series converges by the Alternating Series Test,
but
1
lim arctan n n→ 1
n1
0
2
Therefore, the series diverges by the nth-Term Test.
n0
diverges by a limit comparison to the divergent harmonic
series,
1
n.
n1
cos nn 1 1, therefore the series
lim
1n
converges conditionally.
n→ 61.
1n
cos n
2
n
n2
n1
n1
1
n
n1
2
62.
1n1
sin2n 12
n
n
n1
n1
The given series converges by the Alternating Series Test,
but
is a convergent p-series.
Therefore, the given series converges absolutely.
n1
1
sin2n 12
n
n1 n
is a divergent p-series. Therefore, the series converges
conditionally.
63. An alternating series is a series whose terms alternate in
sign. See Theorem 9.14.
65. an is absolutely convergent if an converges. an
is conditionally convergent if an diverges, but an
converges.
1n
.
n
an converges and
a converges.
1
But, a diverges.
n
n
66. (b). The partial sums alternate above and below the
horizontal line representing the sum.
67. False. Let an Then
64. S Sn Rn ≤ an1 (Theorem 9.15)
n
68. True. If an converged, then so would an by
Theorem 9.16.
284
Chapter 9
Infinite Series
69. True. S100 1 1
1 1 . . .
2 3
100
70. False. Let
a
1
Since the next term 101
is negative, S100 is an
overestimate of the sum.
n
b
n
1n
.
n
Then both converge by the Alternating Series Test. But,
a b
n n
71.
1
1 n
n
n1
72.
p
1
n
n1
1
If p 0, then
n
1 n
If p < 0, then
n p
n→ 1
np
lim an lim
diverges.
n→ n1
If p > 0, then lim
1
n, which diverges.
Assume that n p 0 so that an 1n p are
defined for all n. For all p,
diverges.
n1
an1 1
0 and
np
n→ an1 1
0
np
1
1
<
< an.
n1p np
Therefore, the series converges for all p.
1
1
<
an.
n 1 p n p
Therefore, the series converges for p > 0.
73. Since
a n
n1
an for all n
converges we have lim an 0. Thus, there must exist an N > 0 such that aN < 1 for all n > N
and it follows that
a
n→ an2 ≤
> N. Hence, by the Comparison Test,
2
n
n1
converges. Let an 1n to see that the converse is false.
74.
1
1n1
converges, but
diverges.
n
n1
n1 n
76. (a)
xn
n1 n
converges absolutely (by comparison) for 1 < x < 1,
since
xn
< xn and
n
x
n
is a convergent geometric series for 1 < x < 1.
(b) When x 1, we have the convergent alternating
series
1n
.
n
n1
When x 1, we have the divergent harmonic series
1n. Therefore,
xn
converges conditionally for x 1.
n1 n
75.
1
n
n1
2
converges, hence so does
1
n.
n1
4
77. (a) No, the series does not satisfy an1 ≤ an for all n. For
example, 19 < 18.
(b) Yes, the series converges.
S2n 1
1 1 . . .
1
n n
2 3
2
3
12 . . . 21 13 . . . 31 n
1
n
1 . . .
1
1
1
n 1 . . . n
2
2
3
3
As n → ,
1
1
3 1
2 .
S2n →
1 12 1 13
2 2
Section 9.5
78. (a) No, the series does not satisfy an1 ≤ an:
1
n1a
n
1
n1
Alternating Series
(b) No, the series diverges because
1
n diverges.
1
1
1
. . . and
8 3 64
1
1
<
.
8 3
79.
10
n
n1
32
10
n
1
32 ,
n1
80.
n
n1
convergent p-series
2
3
5
converges by limit comparison to
convergent p-series
1
.
n2
81. Diverges by nth-Term Test
lim an n→ 82. Converges by limit comparison to
1
.
convergent geometric series
2n
83. Convergent geometric series
85. Convergent geometric series
r 1e or Integral Test
86. Converges (conditionally) by
Alternating Series Test
87. Converges (absolutely) by
Alternating Series Test
88. Diverges by comparison to
Divergent Harmonic Series:
89. The first term of the series is zero,
not one. You cannot regroup series
terms arbitrarily.
90. Rearranging the terms of an
alternating series can change
the sum.
ln n 1
> for n ≥ 3
n
n
91. s 1 S1
r 87 < 1
84. Diverges by nth-Term Test
lim an n→ 1 1 1 . . .
2 3 4
1 1 1 1 1 1
1
1
. . .
3 2 5 7 4 9 11 6
(i) s4n 1 1 1 1 1 1 . . .
1
1
2 3 4 5 6
4n 1 4n
1
1 1 1 1
1
1
1
s . . .
2 2n 2 4 6 8 10
4n 2 4n
Adding: s4n (ii) lim sn s
n→ 1
1 1 1 1 1
1
1
1
s 1 . . .
S3n
2 2n
3 2 5 7 4
4n 3 4n 1 2n
(In fact, s ln 2.)
1
s 0 since s > .
2
S lim S3n s4n n→ Thus, S s.
3
1
1
s s s s
2 2n
2
2
3
2
285
286
Chapter 9
Section 9.6
1.
Infinite Series
The Ratio and Root Tests
n 1! n 1nn 1n 2!
n 2!
n 2!
2.
2k 2!
2k 2!
2k!
2k2k 12k 2!
n 1nn 1
1
2k2k 1
3. Use the Principle of Mathematical Induction. When k 1, the formula is valid since 1 13
5.
. . 2n 1 2n!
2n n!
21!
. Assume that
21 1!
and show that
1
35.
. . 2n 12n 1 2n 2!
.
2n1n 1!
To do this, note that:
1
35.
. . 2n 12n 1 1
35.
. . 2n 12n 1
2n!
2n 1 Induction hypothesis
2n n!
2n!2n 1 2n 2
2n 1
2n n!
2n!2n 12n 2
2n1n!n 1
2n 2!
2n1n 1!
The formula is valid for all n ≥ 1.
4. Use the Principle of Mathematical Induction. When k 3, the formula is valid since
1
2n n!2n 32n 1
.
.
.
135
2n 5
2n!
and show that
1
2n1n 1!2n 12n 1
.
.
.
.
135
2n 52n 3
2n 2!
To do this, note that:
13
5.
1
. . 2n 52n 3 1
The formula is valid for all n ≥ 3.
1
. . 2n 5
2n 3
2n n!2n 32n 1
2n!
2n 3
2n n!2n 1
2n!
2n 2n 1n!2n 12n 1
2n!2n 12n 2
2n1n 1!2n 12n 1
2n 2!
35.
1
1
2n 12n 2
2n 12n 2
1 233!35
1. Assume that
1
6!
Section 9.6
5.
3
4
n
n
1
n1
8.
34 2169 . . .
6.
The Ratio and Root Tests
3
1
3
9 1
4 n! 4 16 2 . . .
n
7.
n1
3
S1 , S2 1.875
4
3
S1 , S2 1.03
4
Matches (d).
Matches (c).
1n1 4 4
4
. . .
(2n!
2 24
n1
9.
5n 3
4n
n
n1
33
3n1
9 . . .
n!
2
n1
S1 9
Matches (f).
8
4
2
7
2
. . .
10.
4e
n
4
n0
S1 2
S1 2, S2 3.31
S1 4
Matches (b).
Matches (a).
Matches (e).
11. (a) Ratio Test: lim
n→ (b)
(c)
287
an1
n 125
8n1
n1
lim
lim
n→
n→ an
n25
8n
n
n
5
10
15
20
25
Sn
9.2104
16.7598
18.8016
19.1878
19.2491
2
4 . . .
e
5 5
< 1. Converges
8 8
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.
12. (a) Ratio Test: lim
n→ (b)
(c)
an1
an
n 12 1
n 1!
n2 2n 2
lim
lim
2
n→ n→
n 1
n2 1
n!
n
5
10
15
20
25
Sn
7.0917
7.1548
7.1548
7.1548
7.1548
n 1 1 0 < 1. Converges
(d) The sum is approximately 7.15485
10
(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.
0
11
0
13.
n!
3
n0
lim
n→ 14.
n
an1
n 1!
lim
n→ an
3n1
lim
n→ 3n
n!
n1
3
Therefore, by the Ratio Test, the series diverges.
3n
n!
n0
lim
n→ an1
3n1
lim
n→ n 1!
an
lim
n→ n!
3n
3
0
n1
Therefore, by the Ratio Test, the series converges.
288
15.
Chapter 9
n 4 3
n1
Infinite Series
n
16.
an1
n 13
4n1
lim
lim
n→ an
n3
4n
n→ lim
n→ 3n 1
3
4n
4
lim
lim
n→ n→ 2n
n
n1 1
n→ 2n
2
n1
lim
n→ 2n
lim
n→ lim
n→ n2
2n
2n2
2
n 12
n 13
1
2n3
2
Therefore, by the Ratio Test, the series converges.
20.
2
an1
2n1
lim
n→ n 12
an
an 1
n 13
2n1
lim
n→ an
n3
2n
n→ n
n1
n
lim
Therefore, by the Ratio Test, the series converges.
3n 1 3
2n
2
n3
2
lim
19.
2n
n3n
Therefore, by the Ratio Test, the series diverges.
18.
an1
n1
lim
n→ 2n1
an
n→ n
n1
n
an1
n 13n1
lim
n→ an
2n1
lim
n
2
3
n1
Therefore, by the Ratio Test, the series converges.
17.
n 2 1n1n 2
nn 1
n1
an1 lim
n→ Therefore, by the Ratio Test, the series diverges.
n3
n2
≤
an
n 1n 2
nn 1
n2
0
nn 1
Therefore, by Theorem 9.14, the series converges.
Note: The Ratio Test is inconclusive since lim
n→ The series converges conditionally.
21.
1n 2n
n!
n0
lim
n→ 22.
an1
2n1
lim
n→ n 1!
an
lim
n→ n!
2n
2
0
n1
lim
n→ n→ n
n→ 3
n3n
n!
lim
Therefore, by the Ratio Test, the series diverges.
n2
3
2n
3n2
2n2 2n 1
3
> 1
2
Therefore, by the Ratio Test, the series diverges.
24.
n
an1
3
2n1
lim 2
n→ n 2n 1
an
lim
n!
an1
n 1!
lim
n→ n 13n1
an
lim
n3
n1
n→ Therefore, by the Ratio Test, the series converges.
23.
1n13
2n
n2
n1
2n!
5
n1 n
lim
n→ an1
2n 2!
lim
n→ n 15
an
n5
2n!
2n 22n 1n5
n→ n 15
lim
Therefore, by the Ratio Test, the series diverges.
an1
1.
an
Section 9.6
25.
4n
n!
n0
lim
n→ n!
4n
4
0
n1
nn
n!
26.
a
4n1
lim n1 lim
n→ n→ n 1!
an
n1
The Ratio and Root Tests
an1
n 1n1
lim
n→ an
n 1!
lim
n→ lim
n 1n 1
n 1n!nn
lim
n n 1
n→ Therefore, by the Ratio Test, the series converges.
n!
nn
n→ n n!
n
e > 1
Therefore, by the Ratio Test, the series diverges.
27.
3n
n 1
n
n0
lim
n→ an1
3n1
lim
n→ n 2n1
an
n 1n
3n 1n
3
n1
lim
lim
n→ n 2n1
n→ n 2 n 2
3n
n→ n
0
1e 0
nn 12 , let y lim nn 12 . Then,
n
To find lim
n
n→ ln y lim n ln
n→ ln y lim
n→ 1
n 2 0
nn 12 lim lnn 1
n
0
n→ 1
n 1 1
n 2
1 by L’Hôpital’s Rule
1
n2
1
y e1 .
e
Therefore, by the Ratio Test, the series converges.
28.
n!2
3n!
n0
lim
n→ 3
29.
an1
n 1!
lim
n→ 3n 3!
an
lim
n→ 3n!
2
n!2
n0
lim
n
n→ n 12
0
3n 33n 23n 1
4n
1
an1
4n1
lim n1
n→ 3
an
1
3n 1
4n
lim
43n 1
3n1 1
lim
41 1
3n 4
3 1
3n
3
n→ Therefore, by the Ratio Test, the series converges.
n→ Therefore, by the Ratio Test, the series diverges.
30.
1n 24n
2n 1!
n0
lim
n→ an1
24n4
lim
n→ 2n 3!
an
24
2n 1!
lim
0
4n
n→ 2n 32n 2
2
Therefore, by the Ratio Test, the series converges.
31.
1n1n!
1 3 5 . . . 2n 1
n0
lim
n→ an1
lim
n→ 1
an
35.
n 1!
. . 2n 12n 3
1
35.
. . 2n 1
n!
Therefore, by the Ratio Test, the series converges.
Note: The first few terms of this series are 1 1
1
3
2!
1
35
3!
1
lim
n→ 357
1
n1
2n 3 2
. . ..
289
290
32.
Chapter 9
Infinite Series
1n 2 4 6 . . . 2n
2 5 8 . . . 3n 1
n1
an1
2 4 . . . 2n2n 2
lim
n→ 2 5 . . . 3n 13n 2
an
lim
n→ 2
5 . . . 3n 1
2 4 . . . 2n
Therefore, by the Ratio Test, the series converges.
2 24 2
Note: The first few terms of this series are 2 25 2
33.
1
n
lim
n→ an1
1
lim
n→ n 13
2
an
lim
n→ n n 1
3
2
n3
2
1
n→ n→ 37.
2n 1
n
n4
n
lim
n→ n 1
1
4
1
n→ 38.
n→ n→ 2n n 1
n
2n 1
n→ n
n→ 4n 3
2n 1
n 2n 1
n
n
2n
2
n1
n
1n
ln n
n→ n a
lim n lim
n→ n
1n
ln nn
1
lim
0
n→ ln n
Therefore, by the Root Test, the series converges.
n→ lim
n→ 4n2n 31
n
n
4n 3
2
2n 1
Since 2 > 1, the series diverges.
42.
n
n a
lim n lim
n
Since 2 > 1, the series diverges.
n1
2nn 11 2
n→ np
n
lim
n→ n 1
1
Therefore, by the Root Test, the series diverges.
40.
lim
n2
n
n→ 2nn 11
2n
lim
n
n→ 41.
an1
1
lim
n→ n 1 p
an
n
1
2n 1 2
n
lim an lim
n→ p
n 1
n→ n2
1
1
n a
lim n lim
n
Therefore, by the Root Test, the series converges.
n1
n1
lim
39.
lim
n
n a
lim n lim
n→ 1
2
n1
2
1
n
n1
n1
n n 1
Ratio Test is inconclusive.
36.
an1
1
lim
n→ n 11
2
an
n→ 4
an1
1
lim
n→ n 14
an
1
1
2
lim
1
lim
lim
2n 2 2
3n 2 3
. ..
1
n
n1
n→ n
n1
Ratio Test is inconclusive.
35.
lim
46.
58
34.
3
2
n1
3n
2n 1
3n
n1
lim
n→ n
lim
an n→
lim
n→ 2n3n
1
n
3n
2n3n 1 32
3
3
27
8
Therefore, by the Root Test, the series diverges.
p
1
Section 9.6
43.
The Ratio and Root Tests
291
2n 1
n
n
n1
n a
lim n lim
n→ n→ n 2
n n 1
lim 2 n n 1n n→
n n, let y lim n n . Then
To find lim n→ n→ n n lim
ln y lim ln n→ n→ 1
ln n
1
n
ln n lim
lim
0.
n→ n
n→ 1
n
n n 1 21 1 3. Therefore, by the Root Test, the series diverges.
Thus, ln y 0, so y e0 1 and lim 2
n→ 44.
e
n
45.
n0
n
n
n1
e1 1e
n a
lim n lim
n→ 4
n
n→ n
lim an lim
n→ n
n→ 4n
n
n
n1
n 1
n→ 4
4
Therefore, by the Root Test, the series converges.
lim
1
Since 4 < 1, the series converges.
Note: You can use L’Hôpital’s Rule to show lim n1
n 1.
n→ Let y n1
n
lim
n→ ln n
⇒ ln y n
ln n
1
n
lim
0.
n→ 1
n
Hence, ln y → 0 ⇒ y n1
n → 1.
46.
500
n
n
47.
n1
500n n
n→ ln n
49.
lnnn
n
n
lim an lim
n
n→ n!n
n1
n 2
n!n
n n1
nn!
lim
n!
n2 n→ n→ n
2 n
The series diverges.
n→ lnnn
n
n
n1
n
0
n→ ln n
Since 0 < 1, the series converges by the Root Test.
51.
2 n
n
lim an lim
n→ 2
lim
By the Root Test, the series converges.
1n n1 0 0 0 < 1
n
ln n
0 < 1
n→ n
n 2
n
lim
50.
n
n
ln n
n2
n→ 1n n1 Hence, the series converges.
n
n a
lim n lim
n→ n→ n1
n→ lim
The series diverges.
n
2
n
500
n→ n
1
n a
lim n lim
n
n→ lim
48.
1
n1
n a
lim n lim
n→ n n 1n1 5
n
n1
an1 lim
5
5
< an
n1
n
n→ 5
0
n
Therefore, by the Alternating Series Test, the series
converges (conditional convergence).
292
52.
Chapter 9
5
Infinite Series
1
n 5 n
n1
53.
n1
4
3
n1
This is the divergent harmonic series.
54.
1
nn 3 n
n1
3
2
This is convergent p-series.
n
55.
2n
n1
n1
n1
Since 4 < 1,this is convergent geometric series.
lim
n→ 2n
20
n1
This diverges by the nth-Term Test for Divergence.
56.
2n
2
n1
n
1
57.
1n 3n 32
1
1n 3n2
3
n
2
2n
2
n1
n1
n1 9
n
3
Since r 2 > 1, this is a divergent geometric series.
1
n
2n2 1
n2
lim
> 0
n→ n→ 2n2 1
1
n
2
lim
This series diverges by limit comparison to the divergent
harmonic series
1
n.
n1
58.
10
3n
3
n1
10n 3
n2n
n1
lim
10
3n3
2 10
n→ 1
n3
2
3
10n 3
n2n
10n 3
lim
10
n→ n→ 1
2n
n
Therefore, the series converges by a Limit Comparison
Test with the p-series
Therefore, the series converges by a Limit Comparison
Test with the geometric series
n
n1
60.
59.
4n
lim
n→ 2 .
1
3
2 .
2n
1
2
n1
lim
1
n
n0
61.
n1
cos n
2n
2n
ln 22n
ln 222n
lim
lim
n→ 4n2 1 n→ 8n
8
cos n
1
≤ n
2n
2
Therefore, the series
Therefore, the series diverges by the nth-Term Test for
Divergence.
n1
cos n
2n
converges by comparison with the geometric series
2 .
1
n
n0
62.
1n
n2 n ln n
an1 lim
n→ 1
1
≤
an
n 1 lnn 1
n lnn
1
0
n lnn
Therefore, by the Alternating Series Test, the series
converges.
63.
n7n
n1 n!
lim
n→ an1
n 17n1
lim
n→
an
n 1!
n!
n7n
lim
n→ 7
0
n
Therefore, by the Ratio Test, the series converges.
Section 9.6
64.
lnn
2
n1 n
1n 3n1
n!
n1
lim
n→ n
n1
66.
3
2
an1
3n
lim
n→ n 1!
an
n!
3n1
lim
n→ 3
0
n1
Therefore, by the Ratio Test, the series converges.
Therefore, the series converges by comparison with the
p-series
1
293
65.
1
lnn
≤ 3
2
n2
n
The Ratio and Root Tests
(Absolutely)
.
1n 3n
n2n
n1
lim
n→ an1
3n1
lim
n→ n 12n1
an
3n
3
n2n
lim
n→ 2n 1
3n
2
Therefore, by the Ratio Test, the series diverges.
67.
3n
3 5 7 . . . 2n 1
n1
lim
n→ an1
lim
n→ 3
an
57.
3n1
. . 2n 12n 3
3
57.
3
57.
. . 2n 1
3
lim
0
n→ 2n 3
3n
Therefore, by the Ratio Test, the series converges.
68.
3
n1
lim
n→ 57.
. . 2n 1
18n 2n 1n!
an1
3 5 7 . . . 2n 12n 3
lim
n→ an
18n12n 12n 1n!
1
18n 2n 1n!
2n 32n 1
lim
. . 2n 1 n→
18
2n
1
2n
1
18
Therefore, by the Ratio Test, the series converge.
69. (a) and (c)
70. (b) and (c)
n 15n1
n5n
n!
n 1!
n1
n0
n 14
5
3
n
n0
25
35
45
. . .
2!
3!
4!
2
3
4
n 4
3
n1
12
71. (a) and (b) are the same.
n1
34 334
2
4
34
3
. . .
72. (a) and (b) are the same.
1n
n 12
n2
n1
1
1
1
. . .
2 2 22 3 23
1n1 1
1
1
. . .
n2n
2 2 22 3 23
n1
73. Replace n with n 1.
n1
n
n n1
4
n1
n0 4
74. Replace n with n 2.
2
2
n
2
!
n2
n0 n!
n
n2
75. Since
310
1.59 105,
2 10!
10
use 9 terms.
3k
0.7769
k
k1 2 k!
9
294
76.
Chapter 9
Infinite Series
3k
3k 2k k!
1 3 5 . . . 2k 1 2k!2k 1
k0
an1
4n 1
3n 2an
lim
n→ an
an
n→ k0
77. lim
6 k!
4n 1 4
lim
> 1
n→ 3n 2
3
k
2k 1!
k0
0.40967
The series diverges by the Ratio Test.
(See Exercise 3 and use 10 terms, k 9.)
78. lim
an1
2n 1
5n 4an
lim
n→ an
an
n→ 2n 1 2
< 1
5n 4 5
lim
n→ n→ an1
cos n 1
nan
lim
n→ an
an
n→ lim
n→ an1
sin n 1
nan
lim
n→ an
an
lim
n→ The series converges by the Ratio Test.
80. lim
79. lim
sin n 1
0 < 1
n
The series converges by the Ratio Test.
81. lim
n→ cos n 1
0 < 1
n
an1
1 1
nan
lim
n→ an
an
lim 1 n→ The series converges by the Ratio Test.
1
1
n
The Ratio Test is inconclusive.
But, lim an 0, so the series diverges.
n→ 82. The series diverges because lim an 0.
83. lim
n→ a1 1
4
1
2
1
3
1
a2 4
1
a3 2
n→ lim
1
2
an1
1
lim
n→ an
n→ 1 2 . . . nn 1
3 . . . 2n 12n 1
1 2. . .n
1 3 . . . 2n 1
n1
1
< 1
2n 1 2
The series converges by the Ratio Test.
0.7937
In general, an1 > an > 0.
84.
n1
3n
n0
85.
n3
n 3 1 lim
n
lim an lim
n→ n
n→ n
n→ n n 1
3
n n 1
y lim Let
n→ ln y lim
n→ n n
ln 1
lim
1
lnn 1
n
lim
lnn 1
1
0.
n
n1
n→ n→ Since ln y 0, y e0 1, so
lim
n→ n 1
3
1
.
3
Therefore, by the Root Test, the series converges.
1
ln n
n
n a
lim n lim
n→ n→ ln1n
n
n
lim
n→ 1
0
ln n
Therefore, by the Root Test, the series converges.
a
1
86. lim n1 lim
n→ an
n→ 1 3 5 . . . 2n 12n 1
2 3 . . . 2n 12n2n 1
1 3 5 . . . 2n 1
1 2 3 . . . 2n 1
2n 1
lim
0 < 1
n→ 2n2n 1
Section 9.6
87. lim
The Ratio and Root Tests
n→ an1
2x
3n1
lim
n→ 2x
3n
an
lim
n→ x
x
3
3
For the series to converge:
The series converges by the Ratio Test.
x
< 1 ⇒ 3 < x < 3.
3
For x 3, the series diverges.
For x 3, the series diverges.
Answer: 3 < x < 3
88. lim
n→ x 4 1
n a
lim
n n→
n
89. lim
n
n→ x1
x1
4
4
lim
n→ lim
n→ x 1 < 1
x1
< 1 ⇒ 4 < x 1 < 4
4
For x 0,
1 , diverges.
For x 3,
n
n0
For x 5,
⇒ 1 < x 1 < 1
1n
, converges.
n
n1
n
1
1n1n
, diverges.
n
n1
n1 n
Answer: 2 < x ≤ 0
n0
Answer: 5 < x < 3
a
91. lim n1 lim
n→ n→ an
lim x 1 x 1
n→ n→ ⇒ 1 < x 1 < 1
21 , diverges.
n
n0
For x 0,
21 , diverges.
n
n0
Answer: 0 < x < 2
92. lim
n→ n!
x
2
x 1 n1
an1
lim
n→ n 1!
an
x 1n n!
lim
n→ x 1 0
n1
The series converges for all x.
93. See Theorem 9.17, page 597.
x
2
n1
n
x
2
The series converges only at x 0.
⇒ 0 < x < 2.
For x 2,
n 1!
lim n 1
For the series to converge,
x 1 < 1
For x 2,
1 , diverges.
2 x 1 n1
a
90. lim n1 lim
n→ n→ 2 x 1 n
an
⇒ 2 < x < 0.
⇒ 5 < x < 3.
n
x 1 x 1
n1
For the series to converge,
For the series to converge,
an1
x 1n1
n 1
lim
n→ an
x n
n
94. See Theorem 9.18.
295
296
Chapter 9
95. No. Let an Infinite Series
1
.
n 10,000
96. One example is
100 n.
1
n1
1
n 10,000 diverges.
The series
n1
97. The series converges absolutely. See Theorem 9.17.
98. Assume that
lim an1
an L > 1 or that lim an1
an .
n→ n→ Then there exists N > 0 such that an1
an > 1 for all n > N. Therefore,
an1 > an,
n > N ⇒ lim an 0 ⇒
n→ a
n
diverges.
99. First, let
Second, let
n a
lim n r < 1
n→ and choose R such that 0 ≤ r < R < 1. There must
n
exist some N > 0 such that an < R for all n > N.
n
Thus, for n > N an < R and since the geometric
series
n a
lim n r > R > 1.
n→ a
R
n
n
n a
Then there must exist some M > 0 such that n > R
for infinitely many n > M. Thus, for infinitely many n > M,
n
we have an > R > 1 which implies that lim an 0
n→ which in turn implies that
diverges.
n1
n0
converges, we can apply the Comparison Test to
conclude that
a n
n1
converges which in turn implies that
a
n
converges.
n1
100.
1
, p-series
np
lim
n
n1
n→ n
lim an n→
101. Ratio Test:
1
1
lim
1
n p n→ n p
n
Hence, the Root Test is inconclusive.
Note: lim n p
n 1 because if y n p
n, then
ln y lim
n→ Root Test:
n a
lim n lim
n→ n→ p
p
ln n and ln n → 0 as n → .
n
n
n→ Hence y → 1 as n → .
an1
nln np
lim
1, inconclusive.
n→ n 1lnn 1p
an
n→ nln1n
n
p
lim
n→ 1
n1
nln n p
n
lim n1
n 1. Furthermore, let y ln np
n ⇒
ln y p
p ln ln n
ln ln n. lim ln y lim
n→ n→ n
n
lim
n→ p
0 ⇒ lim ln n p
n 1.
n→ lnn1
n
Thus, lim
n→ 1
1, inconclusive.
n1
nln n p
n
Section 9.6
102.
n!2
xn!,
The Ratio and Root Tests
x positive integer
n1
n!2
n!
(a) x 1:
n!, diverges
(b) x 2:
2n! converges by the Ratio Test:
n!2
n 1!2 n!2
n 12
1
lim
< 1
n→ 2n 2!
2n! n→ 2n 22n 1 4
lim
(c) x 3:
lim
n→ n!2
3n! converges by the Ratio Test:
n 1!2 n!2
n 12
lim
0 < 1
3n 3! 3n! n→ 3n 33n 23n 1
(d) Use the Ratio Test:
n 1!2 n!2
xn!
lim n 12
n→ xn 1! xn!
n→ xn x!
lim
The cases x 1, 2, 3 were solved above. For x > 3, the limit is 0. Hence, the series converges for all integers x ≥ 2.
103. For n 1, 2, 3, . . . , an ≤ an ≤ an ⇒ Taking limits as k → , k
n
n1
a ≤ a
n
≤
n
n1
k
a ≤ a
n1
n
≤
n1
a ⇒ a
n
n1
n
n1
k
a .
n
n1
≤
a .
n
n1
104. The differentiation test states that if
U
n
n1
is an infinite series with real terms and f x is a real function such that f 1
n Un for all positive integers n and d 2 f
dx 2
exists at x 0, then
U
n
n1
converges absolutely if f 0 f0 0 and diverges otherwise. Below are some examples.
Convergent Series
1
Divergent Series
1
n , f x x
n, f x x
1 cos n, f x 1 cos x
sin n, f x sin x
3
3
1
1
105. Using the Ratio Test,
lim
n→ nn 1! 197 an1
n!
19
lim
n→ n 1n
an
7
n n 197
n 1
n1
lim
n→
lim
n
1 11
n 197
n
n→ 19
7
1
e
< 1
Hence, the series converges.
n1
n
n1
297
298
Chapter 9
Infinite Series
106. We first prove Abel’s Summation Theorem:
If the partial sums of
Let Sk a
n
a b
are bounded and if bn decreases to zero, then
n n
converges.
k
a . Let M be a bound for S .
i
k
i1
a1b1 a2b2 . . . anbn S1b1 S2 S1b2 . . . Sn Sn1bn
S1b1 b2 S2b2 b3 . . . Sn1bn1 bn Snbn
n1
S b b
i
i
Snbn
i1
i1
is absolutely convergent because Sibi bi1 ≤ Mbi bi1 and
S b b
The series
i
i
i1
i1
b b
i
converges to b1.
i1
i1
Also, lim Snbn 0 because Sn bounded and bn → 0. Thus,
n→ 1
i i
i1
1
1
2. y 8 x 4 2 x 2 1
y-axis symmetry
Three relative extrema
Matches (c)
Parabola
Matches (d)
1
4. y e123 x 13 x 1 1
3. y e12x 1 1
Cubic
Matches (b)
Linear
Matches (a)
f 1 4
4x12
fx 2x32
f 1 2
P1x f 1 f1x 1
6. f x 4
3
x
f8 1
12
P1x f 8 f8x 8
2 P1x 2x 6
P1x 10
f 8 2
4x13
4
fx x43
3
4 2x 1
P1
n→
Taylor Polynomials and Approximations
1. y 2 x 2 1
4
n
a b converges.
lim
1
to finish the problem.
n
Section 9.7
x
n n
n1
Now let bn 5. f x a b
(1, 4)
1
x 8
12
8
1
x
12
3
8
f
−2
6
−2
P1 is called the first degree Taylor polynomial for f at c.
(8, 2)
−4
14
−4
P1 is called the first degree Taylor polynomial for f at c.
Section 9.7
f
4 2
f
4 2
7. f x sec x
fx sec x tan x
Taylor Polynomials and Approximations
fx sec2 x
4 f 4 x 4 P1x f
P1x 2 2 x 4
P1 f
f
4 2
12 x
P1x 2x 1 f
−
4 1
4 f4 x 4 5
( π4 , 2)
f
8. f x tan x
299
4
2
3
P1
4
2
−
2
−1
2
P1 is called the first degree Taylor polynomial for f at c.
−3
P1 is called the first degree Taylor polynomial for f at c.
9. f x 4
x
f 1 4
4x12
10
P2
fx 2x32
f1 2
f x 3x52
f 1 3
(1, 4)
f
−2
f 1
x 12
2
P2 f 1 f1x 1 6
−2
3
4 2x 1 x 12
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
P2x
7.5
4.46
4.215
4.0
3.815
3.66
3.5
f
4 2
fx sec x tan x
f
4 2
f x sec3 x sec x tan2 x
f
10. f x sec x
P2x f
4
4 3
2
−3
4 f4 x 4 f 24 x 4 P2x 2 2 x 3
2 x 4
2
4
3
0
2
2
0.585
0.685
4
0.885
0.985
1.785
f x
1.8270 1.1995
1.2913
1.4142
1.5791
1.8088
4.7043
P2x
15.5414
1.2936
1.4142
1.5761
1.7810
4.9475
x
2.15
1.2160
300
11.
Chapter 9
Infinite Series
f x cos x
(b)
P2x 1 1 2
2x
P4x 1 1 2
2x
P6x 1 1 2
2x
(a)
1 4
24 x
1 4
24 x
P2x x
f x cos x
P2 x 1
f 0 P2 0 1
1 6
720 x
f x sin x
P4 x x
f 4x cos x
2
P6
−3
fx sin x
f 40 1 P440
f 5x sin x
P65x x
P4
3
f
P44x 1
P2
f 6x cos x
−2
P6x 1
f 60 1 P660
(c) In general, f n0 Pnn0 for all n.
12. f x x 2ex, f 0 0
(a)
fx x 2 2xex
f x f0 0
ex
f 0 2
fx x 2 6x 6ex
f0 6
x2
4x 2
f 4x x 2 8x 12ex
P3x x 2 3!
13.
f x ex
f
4!
3
P3
−1
x 2 x3 f 0 2 P20
f0 6 P30
x 2 x3
P4x x 2 x3 P2
−3
(c)
12x 4
2
P4
f 40 12
2x 2
x2
P2x 2!
6x3
(b)
f 40 12 P440
x4
2
f 0 1
(d) f n0 Pnn0
14.
f x ex
f 0 1
f0 1
fx ex
f0 1
fx f x ex
f 0 1
f x ex
f 0 1
fx ex
f0 1
f x ex
f0 1
P3x f 0 f0x 1x
2
f 4x
f 0 2 f0 3
x x
2!
3!
ex
f 40 1
e
x
f 5x ex
x
2
6
P5x f 0 f0x 15.
f 50 1
x3
f0 2 f0 3 f 40 4
x x x
2!
3!
4!
f 50 5
x 2 x3
x4
x5
x 1x 5!
2
6
24 120
f x e3x
f 0 1
f0 2
fx 3e3x
f0 3
f x 4e2x
f 0 4
f x 9e3x
f 0 9
fx 8e2x
f0 8
fx 27e
f0 27
f 4x 162x
f 40 16
f x e2x
f 0 1
fx 2e2x
P4x 1 2x 4 2
8
16 4
x x3 x
2!
3!
4!
1 2x 2x 2 4 3 2 4
x x
3
3
16.
3x
f 4x 81e3x f 40 81
P4x 1 3x 1 3x 9 2 27 3 81 4
x x x
2!
3!
4!
9 2 9 3 27 4
x x x
2
2
8
Section 9.7
17.
f x sin x
f 0 0
f0 1
fx cos x
f0 f x sin x
f 0 0
f x 2 sin x
f 0 0
fx cos x
f0 1
fx cos x
f0 3
f x sin x
f 0 0
fx cos x
3
f 40 0
f 5x cos x
f 50 1
P5x 0 1x 0 2 1 3
0
1
x x x 4 x5
2!
3!
4!
5!
fx xex
f x xex
fx xex
xex
f 0 0
20.
f x x 2ex
f 0 0
f0 1
fx 2xe
2ex
f 0 2
f x 2e
3ex
f0 3
fx 6ex 6xex x 2ex
4ex
f 40
x
4
f 0 1
1
x 12
2
f x x 13
f0 1
f 0 2
fx 6
x 14
f 4x 24
x 15
f0 0
x e
2 x
4xe
x
f 0 2
xe
2 x
f 4x 12ex 8xex x 2ex
f0 6
f 40 24
2 2 6 3 24 4
x x x
2!
3!
4!
f0 6
f 40 12
2 2 6 3 12 4
x x x
2!
3!
4!
1
x 2 x3 x 4
2
P4x 0 0x 1 3 1 4
x x
2
6
1
x1
P4x 1 x x
2 2
3
4
x x3 x 4
2!
3!
4!
x x2 fx 3 3
0 2 3 3
x x x
x
2!
3!
6
ex
P4x 0 x f x P3x 0 x 1 3
1 5
x x
6
120
f x xex
f 4x
21.
18.
f 4x sin x
x
19.
Taylor Polynomials and Approximations
22.
f x x
x11
x1
x1
f 0 0
1 x 11
fx x 12
f0 1
f x 2x 13
f 0 2
fx 6x 14
f0 6
f 4x
24x 1
f 40 24
5
2
6
24
P4x 0 1x x2 x3 x4
2
6
24
x x2 x3 x4
1 x x 2 x3 x 4
23. f x sec x
f 0 1
24. f x tan x
f 0 0
fx sec x tan x
f0 0
fx sec2 x
f x sec3 x sec x tan2 x
f 0 1
f x 2 sec x tan x
P2x 1 0x 1 2
1
x 1 x2
2!
2
f0 1
f 0 0
2
fx 4
sec2
x
tan2
x2
sec4
x
f0 2
2
1
P3x 0 1x 0 x3 x x3
6
3
301
302
25.
Chapter 9
f x 1
x
f 1 1
fx f x 1
x2
2
x3
fx f 4x Infinite Series
6
x4
f 2 f1 1
fx 4x3
f2 f 1 2
f x 12x4
f 2 f 41 24
P4x 1 x 1 2
6
x 12 x 13
2!
3!
1
2
3
4
3
2
15
f 4x 240x6
f 4x 4
1 1
3
1
P4x x 2 x 22 x 23
2 2
8
4
fx 48x5
f1 6
24
x5
1
2
f x 2x2
26.
24
x 14
4!
fx 5
x 24
32
1 x 1 x 12 x 13 x 14
27.
f x x
fx f 1 1
1
2
x
f1 1
2
f x 1
4x
x
f 1 fx 3
f1 8x 2
x
f 4x 15
16x3
x
29.
f 41 1
x
2
x3
f 4x f 1 0
f1 1
1
f x 2
x
fx 1
fx x23
3
f8 2
f x x53
9
f 8 fx 15
16
f 8 2
10 83
x
27
P3x 2 f8 1
12
1
144
10
27
1
5
28 3456
1
1
5
x 8 x 82 x 83
12
288
20,736
1
5
x 13 x 14
16
128
f x ln x
fx 1
4
3
8
1
1
P4x 1 x 1 x 12
2
8
28. f x x13
f 1 1
f1 2
6
x4
f 41 6
1
P4x 0 x 1 x 12
2
1
1
x 13 x 14
3
4
30. f x x 2 cos x
fx cos x x 2 sin x
f 2
f 2
f x 2 cos x 4x sin x x 2 cos x f 2 2
P2x 2 2x 2 2
x 2
2
Section 9.7
f x tan x
31.
Taylor Polynomials and Approximations
(a) n 3, c 0
fx sec2 x
P3x 0 x f x 2
sec2
x tan x
fx 4
sec2
x tan2 x 2 sec4 x
(b) n 3, c f 4x 8 sec2 x tan3 x 16 sec4 x tan x
0 2
2
1
x x3 x x3
2!
3!
3
4
4
x
4
2!
4
2 x
4
4
2
Q3x 1 2 x f 5x 16 sec2 x tan4 x 88 sec4 x tan2 x 16 sec6 x
6
12 x
−
P5
2 P3
2
f
303
2
16
x
3!
4
8
x
3
4
Q3
−6
32.
f x 1
x2 1
2
P4
2x
fx 2
x 12
f x 23x 2 1
x 2 13
fx 24x1 x 2
x 2 14
f 4x f
−3
3
Q4
P2
−2
245x 4 10x 2 1
x 2 15
(a) n 4, c 0
P4x 1 0x 2 2
0
24 4
x x3 x 1 x2 x 4
2!
3!
4!
(b) n 4, c 1
Q4x 33.
1 1
1
1
12
0
3
1
1
x 1 x 12 x 13 x 14 x 1 x 12 x 14
2
2
2!
3!
4!
2 2
4
8
f x sin x
P1x x
P3x x 16 x3
1
1
P5x x 6 x3 120 x5
(a)
x
0.00
0.25
0.50
0.75
1.00
sin x
0.0000
0.2474
0.4794
0.6816
0.8415
P1x
0.0000
0.2500
0.5000
0.7500
1.0000
P3x
0.0000
0.2474
0.4792
0.6797
0.8333
P5x
0.0000
0.2474
0.4794
0.6817
0.8417
(c) As the distance increases, the accuracy decreases.
(b)
3
P1
P3
f
−2
2
P5
−3
3
3
304
34.
Chapter 9
Infinite Series
f x ln x, c 1
P1x x 1
P2x x 1 12x 12
P4x x 1 12x 12 13x 13 14x 14
(a)
(b)
2
f
P1
x
1.00
1.25
1.50
1.75
2.00
ln x
0
0.2231
0.4055
0.5596
0.6931
P1x
0
0.2500
0.5000
0.7500
1.0000
P2x
0
0.2188
0.375
0.4688
0.5000
P4x
0
0.2230
0.4010
0.5303
0.5833
−1
5
P2
P4
−2
(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) P3x x (b)
x3
6
(c)
y
π
2
x
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
P3x
0.820
0.521
0.253
0
0.253
0.521
0.820
f
x
−1
1
P3
−
36. (a)
f x arctan x
P3x x (b)
y
(c)
π
2
x3
3
π
4
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
P3x
0.6094
0.4583
0.2448
0
0.2448
0.4583
0.6094
37. f x cos x
π
2
P8
x
−1
1
2
−π
4
f
1
−π
2
38. f x arctan x
y
P4
P3
y
P7
P5 P1
6
2
4
f(x) = cos x
2
1
x
−6
6
f (x) = arctan x
8
x
−3 −2
−4
1
3
−6
P6 P2
−2
P9 P13
39. f x) lnx2 1
y
40. f x 4xex 4
P11 P3
2
P6
y
P2
3
2
f(x) = ln (x 2 + 1)
1
2
y = 4xe(−x /4)
4
2
x
−4 −3 −2
2
−3
P8
P4
3
4
−4
x
4
Section 9.7
41. f x ex 1 x f
x 2 x3
2
6
f x x 2ex x 2 x3 42.
12 0.6042
Taylor Polynomials and Approximations
f
1 4
x
2
15 0.0328
f x ln x x 1 12 x 12 13 x 13 14 x 14
43.
f 1.2 0.1823
f x x 2 cos x 2 2x 44.
f
2
2
x 2
2
78 6.7954
45. f x cos x; f 5x sin x ⇒ Max on 0, 0.3 is 1.
1
0.35 2.025 105
5!
Note: you could use R5 x: f 6x cos x, max on 0, 0.3 is 1.
R4x ≤
R5 x ≤
1
0.36 1.0125 10 6
6!
Exact error: 0.000001 1.0 10 6
46. f x ex; f 6x ex ⇒ Max on 0, 1 is e1.
R5x ≤
e1 6
1 0.00378 3.78 103
6!
47. f x arcsin x; f 4x R3x ≤
x6x 2 9
⇒ Max on 0, 0.4 is f 40.4 7.3340.
1 x 272
7.3340
0.44 0.00782 7.82 103. The exact error is 8.5 104. Note: You could use R4.
4!
48. f x arctan x; f 4x 24xx 2 1
1 x 24
49. gx sin x
gn1x
⇒Max on 0, 0.4 is f 40.4 22.3672.
R3x ≤
Rnx ≤
22.3672
0.44 0.0239
4!
≤ 1 for all x.
1
0.3n1 < 0.001
n 1!
By trial and error, n 3.
50. f x cos x
51.
f n1x ≤ 1 for all x and all n.
Rnx ≤
f n1zx n1
n 1!
0.1
< 0.001
n 1!
n1
By trial and error, n 2.
f x ex
f n1x ex
Max on 0, 0.6 is e0.6 1.8221.
Rn ≤
1.8221
0.6n1 < 0.001
n 1!
By trial and error, n 5.
305
306
Chapter 9
Infinite Series
52. f x e x
53. f x lnx 1
f n1x e x
f n1x The maximum value on 0, 0.3 is e0.3 1.3499.
Rn
Rn ≤
1.3499
≤
0.3n1 < 0.001
n 1!
f x cos x2
55. f x ex, f (1.3
gx cos x 1 f x g
0.5n1
n!
< 0.0001
0.5n1 n 1!
n1
By trial and error, n 9. (See Example 9.) Using 9 terms,
ln1.5 0.4055.
By trial and error, n 3.
54.
1n n!
⇒ Max on 0, 0.5 is n!.
x 1n1
fx ex
x2 x 4 x6 . . .
2!
4!
6!
f n1x n1ex ≤ n1 on 0, 1.3
x2
Rn
x 22 x 24 x 26 . . .
1
2!
4!
6!
≤
1.3n1 < 0.0001
n 1!
n1
By trial and error, n 16.
2x 4 4x8 6x12 . . .
1
2!
4!
6!
2
4
6
0.64 0.68 0.612 . . .
2!
4!
6!
f 0.6 1 Since this is an alternating series,
Rn ≤ an1 2n
0.64n < 0.0001.
2n!
By trial and error, n 4. Using 4 terms f 0.6 0.4257.
56. f x ex
57.
fx ex
f
n1x
Rn
≤
1
n1ex
≤ 1 on 0, 1
1
1n1 ≤ 0.0001
n 1!
By trial and error, n 7.
f x ex 1 x R3x x 2 x3
, x < 0
2
6
ez 4
x < 0.001
4!
ez x 4 < 0.024
xe z4
x
< 0.3936
<
0.3936
< 0.3936, z < 0
e z4
0.3936 < x < 0
58.
f x sin x x R3x x4
sin z 4
x ≤
< 0.001
4!
4!
x 4 < 0.024
x
x3
3!
< 0.3936
0.3936 < x < 0.3936
59. f x cos x 1 x4
x2
, fifth degree polynomial
2!
4!
f n1x ≤ 1 for all x and all n.
1
R5x ≤ 6! x6 < 0.001
x6
x
< 0.72
< 0.9467
0.9467 < x < 0.9467
Note: Use a graphing utility to graph
y cos x 1 x22 x 424 in the viewing window
0.9467, 0.9467 0.001, 0.001 to verify the
answer.
Section 9.7
Taylor Polynomials and Approximations
4
60. f x e2x 1 2x 2x2 x 3
3
fx 2e2x, f x 4e2x, f x 8e2x, f 4x 16e2x
R3 x f 4z
16e2z 4 2 2z 4
x 04 x e
x < 0.001
4!
24
3
e2zx 4 < 0.0015
0.0015
e
14
x <
2z
0.1970e2z < 0.1970, for z < 0.
Thus 0 < x < 0.1970.
In fact, by graphing f x e2x and y 1 2x 2x 2 43 x 3, you can verify that 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 P2c, fc P2c, and f c P2 c
63. See definition on page 650.
64. See Theorem 9.19, page 654.
65. The accuracy increases as the degree increases (for values
within the interval of convergence).
66.
y
P2
10
f
P1
8
6
4
2
−20
x
P3
−2
10
20
−4
68. (a) P5x x 67. (a) f x e x
P4x 1 x 1 2 1 3
1 4
x x x
2
6
24
gx xe x
1
1
1 5
x
Q5x x x x3 x 4 2
6
24
2
Q5x x P4x
(b) f x sin x
P5x x x2 x 4
2!
4!
This is the Maclaurin polynomial of degree 4 for
gx cos x.
(b) Q6x 1 x2 x 4
x6
for cos x
2
4!
6!
Q6x x x5
x3
3! 5!
Q6x x P5x x 2 x3
x5
P5x
3! 5!
(c) Rx 1 x gx x sin x
(c) gx P5x 1 x3
x5
for f x sin x
3! 5!
x4
x6
3!
5!
sin x 1
x2 x 4
P5x 1 x
x
3!
5!
x3
x4
x2
2! 3!
4!
Rx 1 x x2
x3
2! 3!
The first four terms are the same!
307
308
Chapter 9
Infinite Series
69. (a) Q2x 1 2x 22
32
(b) R2x 1 2x 62
32
(c) No. The polynomial will be linear.
Translations are possible at
x 2 8n.
70. Let f be an odd function and Pn be the nth Maclaurin polynomial for f. Since f is odd, f is even:
fx lim
h→0
f x h f x
f x h f x
f x h f x
lim
lim
fx.
h→0
h→0
h
h
h
Similarly, f is odd, f is even, etc. Therefore, f, f , f 4, etc. are all odd functions, which implies that f 0 f 0 . . . 0.
Hence, in the formula
Pnx f 0 f0x f 0x2 . . .
all the coefficients of the even power of x are zero.
2!
71. Let f be an even function and Pn be the nth Maclaurin polynomial for f. Since f is even, f is odd, f is even, f is odd, etc.
All of the odd derivatives of f are odd and thus, all of the odd powers of x will have coefficients of zero. Pn will only have
terms with even powers of x.
f ic
72. Let Pn x a0 a1x c a2x c2 . . . an x cn where ai .
i!
Pnc a0 f c
For 1 ≤ k ≤ n, Pnkc an k! f k!ck! f
k
k
c.
73. As you move away from x c, the Taylor Polynomial becomes less and less accurate.
Section 9.8
Power Series
2. Centered at 0
1. Centered at 0
5.
1
n
n0
L lim
n→ lim
n→ x
7.
xn
n1
6.
un1
1n1xn1
lim
n→ un
n2
n1
1n xn
n1
x x
n2
4. Centered at 3. Centered at 2
2x
n
n0
L lim
n→ un 1
2xn1
lim
2x
n→ un
2xn
2x < 1 ⇒ R
1
2
< 1⇒R1
2xn
2
n1 n
L lim
n→ lim
n→ 8.
un1
2xn1
lim
n→
un
n 12
2n2x
n 12
2 x < 1⇒R
1
2
2x
n2
2xn
1n xn
2n
n0
L lim
n→ 1
x
2
un1
1n1xn1
lim
n→
un
2n1
1
x < 1⇒R2
2
2n
1n xn
Section 9.8
9.
2x2n
n0 2n!
L lim
n→ 10.
un1
2x2n22n 2!
lim
n→ un
2x2n2n!
lim
n→ 2
x
L lim
n→ un 1
2n 2!x2n2n 1!
lim
n→
un
2n!x2nn!
lim
n→ 2n 22n 1
n 1
x2
The series only converges at x 0. R 0.
n
12.
5
x
n0
Since the series is geometric, it converges only if
x2 < 1 or 2 < x < 2.
is geometric, so it converges if
x5 < 1 ⇒ x < 5 ⇒ 5 < x < 5.
1n x n
n
n1
lim
n→ 14.
un1
1n1xn1
lim
n→
un
n1
lim
n→ n
1n xn
nx
x
n1
Interval: 1 < x < 1
n1
lim
n→ n 1xn
un 1
1n2n 2xn1
lim
n→ un
1nn 1xn
lim
n→ n 2x
x
n1
Interval: 1 < x < 1
1n
When x 1, the alternating series
converges.
n
n1
1
n0
n
n0
13.
2n!x2n
n!
n0
309
2x2
0
2n 22n 1
Thus, the series converges for all x. R .
11.
Power Series
When x 1, the series
1
n1
n 1 diverges.
n0
When x 1, the p-series
1
n diverges.
When x 1, the series
n1
n0
Therefore, the interval of convergence is 1 < x ≤ 1.
15.
Therefore, the interval of convergence is 1 < x < 1.
xn
n!
n0
16.
lim
n→ un1
x
lim
n→ n 1!
un
lim
n→ n!
xn
lim
2n!2
x
3xn
n→ x
0
n1
un1
3xn1
lim
n→ 2n 1!
un
lim
n→ The series converges for all x. Therefore, the interval of
convergence is < x < .
17.
2n!
n0
n1
lim
n
un1
2n 2!x n1
lim
n→ un
2n1
2n
2n!x n
Therefore, the series converges only for x 0.
lim
n→ 2n!
3xn
3x
0
2n 22n 1
Therefore, the interval of convergence is < x <
n0
n→ n 1 diverges.
2n 22n 1 x
2
.
310
18.
Chapter 9
Infinite Series
1n xn
n 1n 2
n0
lim
n→ un1
1n1xn1
lim
n→ n 2n 3
un
Interval: 1 < x < 1
When x 1, the alternating series
n 1n 2
n 1x
x
lim
n→ 1n xn
n3
1n
n 1n 2 converges.
n0
1
1
n 1n 2 converges by limit comparison to n .
When x 1, the series
n0
n1
2
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
19.
1n1x n
4n
n1
Since the series is geometric, it converges only if x4 < 1 or 4 < x < 4.
20.
1n n!x 4n
3n
n0
lim
n→ un1
1n1n 1!x 4n1
lim
n→ un
3n1
3n
1n n!x 4n
R0
lim
n→ n 1x 4
3
Center: x 4
Therefore, the series converges only for x 4.
21.
1n1x 5n
n5n
n1
lim
n→ un1
1n2x 5n1
lim
n→ un
n 15n1
n5n
1n1x 5n
R5
lim
n→ nx 5
1
x5
5n 1
5
Center: x 5
Interval: 5 < x 5 < 5 or 0 < x < 10
When x 0, the p-series
1n1
1
diverges. When x 10, the alternating series
converges.
n
n1 n
n1
Therefore, the interval of convergence is 0 < x ≤ 10.
22.
x 2n1
n 14
n0
lim
n→ n1
un1
x 2n2
lim
n→ n 24n2
un
n 14n1
x 2n1
R4
lim
n→ Center: x 2
Interval: 4 < x 2 < 4 or 2 < x < 6
When x 2, the alternating series
1n1
n 1 converges.
n0
1
When x 6, the series
diverges.
n
1
n0
Therefore, the interval of convergence is 2 ≤ x < 6.
x 2n 1
1
x 2
4n 2
4
Section 9.8
23.
1n1x 1n1
n1
n0
Power Series
lim
n→ un1
1n2x 1n2
lim
n→ un
n2
n1
1n1x 1n1
R1
lim
n→ n 1x 1
x 1
n2
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.
n0
When x 2, the alternating series
1n1
converges.
n0 n 1
Therefore, the interval of convergence is 0 < x ≤ 2.
24.
1n1x 2n
n2n
n1
lim
n→ 25.
an1
1n2x 2n1 1n1x 2n
lim
n→ an
n 12n1
n2n
lim
n→ x2 n
x2
2 n1
2
n1
x3
3
12n
1
1n12n
, diverges.
n
n
n2
n2
n1
n1
n1 n
At x 4,
1n1
1n1 2n
, converges.
n
n2
n
n1
n1
Interval of convergence: 0 < x ≤ 4
26.
1nx2n1
n0 2n 1
lim
n→ un1
1n1x2n3
lim
n→ un
2n 3
2n 1
1nx2n1
R1
lim
n→ Interval: 1 < x < 1
When x 1,
1n
2n 1 converges.
n0
When x 1,
1n1
2n 1
converges.
n0
Therefore, the interval of convergence is 1 ≤ x ≤ 1.
is geometric. It converges if
Interval convergence: 0 < x < 6
At x 0,
n1
x3
< 1 ⇒ x 3 < 3 ⇒ 0 < x < 6.
3
x2
< 1 ⇒ 2 < x 2 < 2 ⇒ 0 < x < 4
2
2n 1 2
x x2
2n 3
311
312
27.
Chapter 9
Infinite Series
n
2xn1
n
1
n1
28.
u
n 12xn
lim n1 lim
n→ n→ un
n2
lim
n→ 2xn 12
2x
nn 2
1
R
2
Interval: n1
n2xn1
1n x 2n
n!
n0
lim
n→ un1
1n1x 2n2
lim
n→ un
n 1!
lim
n→ x2
n1
n!
1n x 2n
0
Therefore, the interval of convergence is < x <
.
1
1
< x <
2
2
1
n
When x , the series
diverges
2
n
1
n1
by the nth Term Test.
1n1n
1
When x , the alternating series
diverges.
2
n1
n1
Therefore, the interval of convergence is 29.
1
1
< x < .
2
2
x 2n1
2n 1!
n0
30.
x 2n3
un1
lim
n→ 2n 3!
un
lim
n→ lim
n→ 2n 1!
x 2n1
2
n1
lim
34.
lim
n→ x2
0
2n 22n 3
n!xn
un1
n 1!xn1
lim
n→ un
2n 2!
lim
n→ .
an1
n 2xn1
n2
x x
lim
lim
n→ n 1xn
n→ n 1
an
Converges if x < 1 ⇒ 1 < x < 1.
At x ± 1, diverges.
Interval of convergence: 1 < x < 1
32.
2
46.
. . 2n
3 5 7 . . . 2n 1 x
n1
lim
n→ 2n1
un1
2 4 . . . 2n2n 2x 2n3
lim
n→ 3 5 7 . . . 2n 12n 3
un
3
2
5 . . . 2n 1
4 . . . 2nx 2n1
R1
When x ± 1, the series diverges by comparing it to
Therefore, the interval of convergence is < x <
2n!
n!xn
n 1x
0
2n 22n 1
. . n 1xn
n 1xn
n!
n1
n→ n1
Therefore, the interval of convergence is < x <
31.
2n!
1
2n 1
n1
which diverges. Therefore, the interval of convergence is 1 < x < 1.
lim
n→ 2n 2x 2
x 2
2n 3
.
Section 9.8
33.
Power Series
1n1 3 7 11 . . . 4n 1x 3n
4n
n1
313
un1
1n2 3
lim
n→ un
lim
n→ lim
n→ 7 11 .
. . 4n 14n 3x 3n1
4n1
1n1
3
7
4n
11
. . . 4n 1x 3n
4n 3x 3
4
R0
Center: x 3
Therefore, the series converges only for x 3.
34.
n!x 1n
1 3 5 . . . 2n 1
n1
lim
n→ an1
lim
n→ 1
an
lim
n→ n 1!x 1n1
n!x 1n
3 5 . . . 2n 12n 1 1 3 5 . . . 2n 1
n 1x 1
1
x 1
2n 1
2
1
Converges if x 1 < 1 ⇒ 2 < x 1 < 2 ⇒ 3 < x < 1.
2
At x 1, an n!2n
2 4 6 . . . 2n
> 1, diverges.
.
.
.
135
2n 1 1 3 5 . . . 2n 1
At x 3, an n!2n
2 4 . . . 2n
1n
, diverges.
.
.
.
13
2n 1
1 3 . . . 2n 1
Interval of convergence: 3 < x < 1
35.
x cn1
cn1
n1
lim
n→ 36.
un1
x cn
lim
n→ un
cn
cn1
x cn1
Rc
1
xc
c
n!kxn
, k is a positive integer.
n0 kn!
lim
n→ lim
n→ Center: x c
Interval: c < x c < c or 0 < x < 2c
When x 0, the series
1
n1
diverges.
Converges if
n1
When x 2c, the series
1 diverges.
n1
Therefore, the interval of convergence is 0 < x < 2c.
37.
k
x
n
n0
n
an1
n 1!kxn1 n!kx
lim
n→ an
kn 1!
kn!
Since the series is geometric, it converges only if xk < 1 or k < x < k..
x
kk
x < 1
kk
n 1 x
k nkk 1 nk . . . 1 nk
k
⇒ R k k.
314
38.
Chapter 9
Infinite Series
1n1x cn
ncn
n1
lim
n→ un1
1n2x cn1
lim
n→ un
n 1cn1
ncn
1n1x cn
Rc
lim
n→ nx c
1
xc
cn 1
c
Center: x c
Interval: c < x c < c or 0 < x < 2c
When x 0, the p-series
n1
1
diverges.
n
1n1
converges. Therefore, the interval of convergence is 0 < x ≤ 2c.
n
n1
When x 2c, the alternating series
39.
kk 1 . . . k n 1x n
n!
n1
lim
n→ un1
kk 1 . . . k n 1k nx n1
lim
n→ un
n 1!
kk 1 .
n!
k nx
lim
x
. . k n 1xn n→
n1
R1
When x ± 1, the series diverges and the interval of convergence is 1 < x < 1.
kk 11 2 . .k .n n 1 ≥ 1
. . .
40.
n!x cn
1 3 5 . . . 2n 1
n1
lim
n→ un1
lim
n→ 1
un
n 1!x cn1
1 3 5 . . . 2n 1
n 1x c
1
lim
x c
.
.
.
n→ 2n 12n 1
n!x c
2n 1
2
35
R2
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 < x < c 2.
1 3n!c5. .2 .2nc 1 1 3 5 .n!.2. 2n 1 1 23 45 .6. . 2n2n 1 > 1
n
41.
xn
x2
x
x n1
n! 1 1 2 . . . n 1!
n0
43.
. . .
2
42.
x 2n1
x 2n1
n0
44.
n1
n 1x n n1
n0
n1
2n 1! 2n 1!
1
1 nx
n
n1
n1
1n1x2n1
1nx2n1
2n 1
n0 2n 1
n1
Replace n with n 1.
45. (a) f x 2 , 2 < x < 2
n
x
(Geometric)
46. (a) f x n0
(b) fx 22
n
x
n1
1n1x 5n
, 0 < x ≤ 10
n5n
n1
(b) fx 1n1x 5n1
, 0 < x < 10
5n
n1
, 2 < x < 2
(c) f x 1n1n 1x 5n2
, 0 < x < 10
5n
n2
, 2 ≤ x < 2
(d)
, 2 < x < 2
n1
(c) f x (d)
2
n
n2
f x dx n1
2
2x 2
x
n
1
2
n0
n2
n1
f x dx 1n1x 5n1
, 0 ≤ x ≤ 10
nn 15n
n1
Section 9.8
1n1x 1n1
,0 < x ≤ 2
n1
n0
47. (a) f x (b) fx 48. (a) f x 1n1x 2n
,1 < x ≤ 3
n
n1
(d)
1n1x 1n2
f x dx ,0 ≤ x ≤ 2
n 1n 2
n1
(d)
3
1
n
1
n0
1 1 . . .
3 9
3 3.1
1
n1
f x dx n 1x 2n2, 1 < x < 3
3
2
1n1x 2n1
,1 ≤ x ≤ 3
nn 1
n1
n
1
n0
2 4 . . .
3 9
S1 1, S2 1.67. Matches (a).
n
52. g2 diverges. Matches (b).
3 alternating. Matches (d).
2
n
n0
n0
53. g
x 2n1, 1 < x < 3
n1
n2
50. g2 S1 1, S2 1.33. Matches (c).
51. g3.1 1
(c) f x 1n1nx 1n1, 0 < x < 2
n1
49. g1 n1
n0
(c) f x 315
(b) fx 1n1x 1n, 0 < x < 2
Power Series
18 218
n
4
n0
1
n
1
n0
1
1
. . ., converges
4 16
S1 1, S2 1.25, S3 1.3125 Matches (b).
81 2 81
54. g n
n0
4 1
n
1
n0
1
1
. . ., converges
4 16
S1 1, S2 0.75, S3 0.8125 Matches (c).
55. g
169 2169 n
n0
S1 1, S2 8 , diverges
9
n
n0
17
Matches (d).
8
57. A series of the form
a x c
n
n
n0
is called a power series centered at c.
56. g
38 238
n
n0
xn
n1
2
n
converges
n0
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.
3
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
diverges for x c > R.
61. Answers will vary.
xn
converges for 1 ≤ x < 1. At x 1, the convergence is conditional because
n1 n
n
4 ,
S1 1, S2 1.75 Matches (a).
59. A single point, an interval, or the entire real line.
converges for 1 ≤ x ≤ 1. At x ± 1, the convergence is absolute.
1
n diverges.
316
Chapter 9
Infinite Series
62. Many answers possible.
x n
x
< 1 ⇒ x < 2
(a)
Geometric:
2
n1 2
(b)
(c)
63. (a) f x n
(d)
1n x 2n1
, < x < n0 2n 1!
x 2n
converges for 2 ≤ x < 6
n4n
n1
1n x 2n
, < x < 2n!
n0
1 x
2n!
n0
n
(b) fx 2n
xn
(b) fx n1
n!
xn
x2
x3
x4
n1
x n1
n1
f 0 1
xn
n0
1n1x 2n1
1n x 2n1
2n 1!
n0 2n 1!
n1
n! 1 x 2! 3! 4! . . .
(c) f x (See Exercise 11)
n 1! n! f x
65. y 1n x2n1
f x
n0 2n 1!
(d) f x sin x and gx cos x
n0
nx n1
gx
n!, < x < 64. (a) f x 1n1x 2n1
1n x 2n1
2n
1
!
2n 1!
n1
n0
(c) gx (See Exercise 29.)
gx Geometric: 2x 1 < 1 ⇒ 1 < x < 0
n1
1nxn
converges for 1 < x ≤ 1
n
n1
2x 1
(d) f x ex
66. y 1n1x 2n2
1n x 2n
2n!
2n 2!
n0
n1
y 1nx 2n
1n 2n 1x2n
2n
1
!
2n!
n0
n0
y 1nx 2n1
1n 2nx2n1
2n
!
n1
n1 2n 1!
y 1x 2n1
1n 2nx 2n1
2n
!
n1
n1 2n 1!
y 1nx 2n2
1n 2n 1x 2n2
2n
1
!
n1
n1 2n 2!
y y 1n1x 2n1
1n x 2n1
0
2n 1!
n1 2n 1!
n1
x 2n1
x 2n1
n0
y y 2n 1! 2n 1!
67. y 68. y 1n1x 2n2
1n x 2n2
0
2n 2!
n1 2n 2!
n1
x 2n
x 2n2
2n! 2n 2!
n0
n1
n1
y 2n 1
n0 2n 1!
n0 2n!
y 2n 2x2n1
x 2n1
2n 2!
n1
n1 2n 1!
y x 2n1
2nx 2n1
y
2n!
n1
n1 2n 1!
y x 2n2
2n 1x 2n2
y
2n 1!
n1
n1 2n 2!
x2n
x 2n
x 2n
n n!
2
n0
y y 0
y y 0
69. y y y xy y 2nx 2n1
n
n1 2 n!
y 2n2n 1x 2n2
2n n!
n1
2nx 2n
x 2n
2n2n 1x 2n2
n n!
n n! n
2
2
n1
n1
n0 2 n!
2n 1x 2n
2n2n 1x 2n2
n
2 n!
2n n!
n1
n0
n0
2n 22n 1x2n 2n 1x2n 2n 1
2n 1
2n1n 1!
2n n!
2n 1x 2n 2n 1 2n 1
0
2n1n 1!
n0
Section 9.8
2
y1
70.
2n
n1
2
y 2n
n1
2
y 2n
n1
n! 3
1n 4n4n 1 x 4n2
1n 4nx 4n2
x 2 2n
.
.
.
. . . 4n 5
n! 3 7 11
4n 1
n2 2 n! 3 7 11
2
n2
2
2n2
n1
71. J0x (a) lim
1n x 4n
. . 4n 1
7 11 .
1n 4nx 4n1
n! 3 7 11 . . . 4n 1
y x 2y x 2 2n
1n 4nx 4n2
1n x 4n2
x2
2n
.
.
.
n! 3 7 11
4n 5 n1 2 n! 3 7 11 . . . 4n 1
1n1 4n 1x4n2
1n1 x 4n2
22n 1
0
n 1! 3 7 11 . . . 4n 1 n1 22n n! 3 7 11 . . . 4n 1 22n 1
1k x 2k
2k
2
k0 2 k!
k→ Power Series
uk1
1k1 x 2k2
lim 2k2
k→ 2
uk
k 1!2
22k k!2
1k x 2k
lim
Therefore, the interval of convergence is < x <
1
J0 (b)
k
k0
2k2k 1x 2k2
2k 22k 1x 2k
1k1
4k k!2
4k1 k 1!2
k0
k
k1
1
k1
k0
(c) P6x 1 4k k!2
1
x 2J0 xJ0 x 2J0 .
x 2k
2kx 2k1
2k 2 x 2k1
1k1 k1
k
2
4 k!
4 k 1!2
k0
k
k1
J0 1x 2
0
22k 12
1
J0 k→ 22k 1 x 2k2
2x 2k2
x 2k2
1k1 k1
1k k 2
k1
4 k 1!k!
4 k 1!k! k0
4 k!
k0
2
1k x2k2
22k 1
1
1
1
4k k!2
4k 1
4k 1
k0
1k x 2k2 4k 2
2
4k 4
0
4k k!2
4k 4
4k 4 4k 4
k0
1
x2
x4
x6
4
64 2304
(d)
0
0
3
−6
1
J0dx 1k x 2k
k
2 dx
k0 4 k!
x
4 1
k! 2k 1
k0
6
k
k
2k1
1
2
0
1k
kk!22k 1
4
k0
−5
1
1
1
0.92
12 320
(integral is approximately 0.9197304101)
72. J1x x
k0
(a) lim
k→ 1k x 2k
1k x 2k1
2k1
k 1! k0 2
k!k 1!
22k1k!
uk1
1k1 x 2k3
lim 2k3
k→ 2
uk
k 1!k 2!
22k1k!k 1!
1k x 2k1
Therefore, the interval of convergence is < x <
—CONTINUED—
.
lim
k→ 1x 2
0
22 k 2k 1
317
318
Chapter 9
Infinite Series
72. —CONTINUED
2
(b) J1x 1k x2k1
k 1!
2k1k!
k0
J1x 1k 2k 1x2k
2k1k!k 1!
k0 2
J1x 1k 2k 12kx 2k1
22k1k!k 1!
k1
x 2J1 xJ1 x 2 1J1 1k2k 1x 2k1
1k 2k 12kx 2k1
2k1
2
k!k 1!
22k1k!k 1!
k1
k0
1k2k 1x 2k1
1k2k 12kx 2k1 x
2k1
2
k!k 1!
2 k1 22k1k!k 1!
k1
1kx 2k12k 12k 2k 1 1
1kx 2k3
2k1
2k1
2
k!k 1!
k!k 1!
k1
k0 2
1kx2k14kk 1
1kx2k3
2k1
2k1
2
k!k 1!
k!k 1!
k1
k0 2
1kx 2k1
1k x2k3
2k1
k 1!k! k0 2
k!k 1!
22k1
1k1x 2k3
1kx 2k3
2k1k!k 1! 2k1k!k 1!
2
k0
2
k0
1kx 2k31 1
0
22k1k!k 1!
k0
x
1 3
1 5
1
x x x7
2 16
384
18,432
J0x 1
n
n0
1k1x 2k1
1kx 2k1
k 1! k0 22k1k!k 1!
2
k0
4
1k1x 2k1
1k12k 1x 2k1
2k2
k 1!k 1! k0 22k1k!k 1!
k0 2
J1x 73. f x k1
(d)
1kx2k1
1kx2k3
x
2k1k!k 1!
2 k1 22k1k!k 1!
2
k0
(c) P7x 2k1k!
k0
1kx 2k3
1k x 2k1
2k1
k 1! k0 2
k!k 1!
2
2k1k!
x 2n
cos x
2n!
−6
Note: J0x J1x
74. f x 2
1
n
n0
− 2
−4
x 2n1
2n 1!
sin x
2
1 x
n n
n0
x
n
Geometric
76. f x n0
1
n
n0
1
for 1 < x < 1
1 x 1 x
x 2n1
arctan x, 1 ≤ x ≤ 1
2n 1
2
3
−2.5
2.5
−
1
0
−2
1
−1
2
−
−2
75. f x 6
2
Section 9.8
77.
2
x
Power Series
319
n
n0
2 (a)
34
n
n0
8
3
n
n0
1
8
1.6
1 38 5
(b)
34
2
n0
n
8 3
n
n0
1.80
1
8
0.7272
1 38 11
1.10
−1
6
0
(c) The alternating series converges more rapidly. The
partial sums of the series of positive terms approach
the sum from below. The partial sums of the alternating
series alternate sides of the horizontal line representing
the sum.
−1
6
0.60
2
N
(d)
3
n
> M
n0
M 10
N
78.
3x
n
n0
100
1000
10,000
9
15
21
4
1 1
converges on , .
3 3
1
(a) x :
6
36
1
n
n0
2
1
n
n0
1
2
1 12
1
(b) x :
6
1
3 6
n
n0
3
1
2 n
n0
1
2
1 12 3
1.5
−1
6
0
−1
(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 horizontal
line y 23.
6
−0.5
3 3 N
(d)
2
n
n0
1 x
n2n
n0
1
0
100
1000
N
3
6
9
10,000
13
n
xn
n0
1
f x dx 10
a
converges for x 2, then we know that it must converge
on 2, 2.
81. True; the radius of convergence is R 1 for both series.
> M
M
n n
converges for x 2 but diverges for x 2.
82. True
n
n0
80. True; if
79. False;
N
2
0
n0
an xn dx an xn1
n0 n 1
n a 1
1
0
n0
n
320
Chapter 9
83. lim
n→ Infinite Series
an1
n 1 p!
n p! n
n 1 px
lim
x n1
x lim
0
n→ n 1!n 1 q!
n→ n 1n 1 q
an
n!n q!
Thus, the series converges for all x : R .
84. (a) gx 1 2x x 2 2x3 x 4 . . .
S2n 1 2x x 2 2x 3 x 4 . . . x 2n 2x 2n1
1 x 2 x 4 . . . x 2n 2x1 x 2 x 4 . . . x 2n
lim S2n n→ x
2x
2n
n0
x
2n
n0
Since each series is geometric, R 1.
(b) For x < 1, gx 85. (a) f x 1
1
1 2x
2x
.
1 x2
1 x2
1 x2
c x ,c
n
n
n3
cn
n0
c0 c1x c2x 2 c0x3 c1x 4 c2x5 c0x6 . . .
S3n c01 x3 . . . x3n c1x1 x3 . . . x3n c2 x 21 x3 . . . x3n
lim S3n c0
n→ x
3n
c1x
n0
x
3n
c2 x 2
n0
x
3n
n0
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→ c x ,
n
n
an1
c x n1
c
cn
lim n1 n
lim n1 x < 1 ⇒ x <
R
n→ n→ an
cn x
cn
cn1
For the series
lim
1
c c1x c2 x 2
1
1
c1x
c2 x 2
0
.
1 x3
1 x3
1 x3
1 x3
c x
n
2n
,
bn1
c x 2n2
c
cn
lim n1 2n
lim n1 x2 < 1 ⇒ x2 <
R ⇒ x < R.
n→ n→ bn
cn x
cn
cn1
87. At x0 R, the series is
c x
n
0
x0 Rn n0
At x0 R, the series is
c R
n
n
which converges.
n0
c R , which diverges.
n
n
n0
Hence, at x0 R, we have
c R
n
n
converges,
n0
⇒ The series converges conditionally.
c R c R
n
n0
n
n
n0
n
diverges.
Section 9.9
Section 9.9
1. (a)
Representation of Functions by Power Series
1
12
a
2 x 1 x2 1 r
Representation of Functions by Power Series
2 2
1 x
n
n0
2. (a) f x xn
2
n0
n1
3. (a)
1
12
a
2 x 1 x2 1 r
2 2 n0
1
x
n
1n xn
2n1
n0
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
x 2 x3
4
8
x3
8
x3
x4
8
16
5 5
4 x
n
n0
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
x 2 x3
4
8
x3
8
x3
x4
8
16
4
45
a
5 x 1 x5 1 r
4xn
5
n0
n1
This series converges on 5, 5.
4
4
4 2
4x3
x
x . . .
5 25
125
625
(b) 5 x ) 4
4
4 x
5
4
x
5
4
4
x x2
5
25
4 2
x
25
4 2
4x3
x 25
125
4x3
125
4x3
4x 4
125 625
4. (a)
1
1
a
1 x 1 x 1 r
x
n0
n
1
n
xn
n0
This series converges on 1, 1.
1 x x 2 x3 . . .
(b) 1 x ) 1
1x
x
x x 2
x2
x 2 x3
x3
x3 x 4
321
322
Chapter 9
Infinite Series
6. Writing f x in the form a1 r, we have
5. Writing f x in the form a1 r, we have
1
13
1
2 x 3 x 5 1 13x 5
4
47
a
4
.
5 x 7 x 2 1 17x 2 1 r
which implies that a 13 and r 13x 5.
Therefore, the power series for f x is given by
1 1
1
ar n x 5
2 x n0
3 3
n0
x 5
3
n
n1 ,
n0
x 5
n
< 3 or 2 < x < 8.
3
ar n 32xn
2x 1 n0
n0
x 2 < 7
1
n
which implies that a 1 and r 23x 2.
Therefore, the power series for f x is given by
< x <
1
.
2
x 2
9. Writing f x in the form a1 r, we have
2nx 2n
,
3n
n0
<
3 1
7
or < x < .
2 2
2
1
1
ar n 2x 5 n0
11
n0
112 x 3
n
1
1
ar n 2x 5 n0
5
n0
5 x
x
<
2
n
2n xn
5
n0
n1
,
5
5
5
or < x < .
2
2
2
2nx 3n
,
11n1
n0
17
5
11
or < x < .
2
2
2
<
11. Writing f x in the form a1 r, we have
3
3
32
a
x 2 2 x 1 12x 1 r
which implies that a 32 and r 12x. Therefore,
the power series for f x is given by
3
1
3
ar n x
x 2 n0
2
n0 2
n
12. Writing f x in the form a1 r, we have
4
4
12
a
3x 2 8 3x 2 1 38x 2 1 r
which implies that a 12 and r 38x 2.
Therefore, the power series for f x is given by
1
3
4
arn x 2
3x 2 n0
8
n0 2
< 2 or 2 < x < 2.
1n x n 3 x n
,
n1
2
2 n0
2
n0
3
x
which implies that a 15 and r 25x. Therefore,
the power series for f x is given by
which implies that a 111 and r 211x 3.
Therefore, the power series for f x is given by
x 3
1
1
15
a
2x 5 5 2x 1 25x 1 r
111
a
1 211x 3 1 r
10. Writing f x in the form a1 r, we have
1
1
2x 5 11 2x 3
or 9 < x < 5
2x , 2x < 1 or 2
4x 2n
.
7n1
n0
n
2
3
ar n x 2 ,
2x 1 n0
3
n0
n0
3
3
1
a
2x 1 3 2x 2 1 23x 2 1 r
which implies that a 3 and r 2x. Therefore, the
power series for f x is given by
n
8. Writing f x in the form a1 r, we have
3
3
a
2x 1 1 2x 1 r
3
4 1
4
ar n x 2
5 x n0
n0 7 7
7. Writing f x in the form a1 r, we have
Therefore, the power series for f x is given by
x 2
<
n
1 3n x 2n
,
2 n0
8n
2
8
14
or < x <
.
3
3
3
Section 9.9
13.
Representation of Functions by Power Series
1
3x
2
1
2
1
1
x 2 x 2 x 2 x 1 2 x 1 x 1 12x 1 x
Writing f x as a sum of two geometric series, we have
3x
1
x
x 2 x 2 n0 2
n
1x
n
2
n0
n→ 14.
un1
1 2n1x n1
lim
n→ un
2n1
1 x n.
n
n0
The interval of convergence is 1 < x < 1 since
lim
1
2n
1 2n xn
lim
n→ 1 2n1x
x.
2 2n1
4x 7
3
2
3
2
32
2
2x2 3x 2 x 2 2x 1 2 x 1 2x 1 12x 1 2x
Writing f x as a sum of two geometric series, we have
3
4x 7
2x 2 3x 2 n0 2
2 x
15.
1
n
22x
n
n0
n0
31n
1
1
1
2n1 xn, x < or < x < .
2n1
2
2
2
2
1
1
1 x2 1 x 1 x
Writing f x as a sum of two geometric series, we have
2
xn xn 1 1n x n 2x 2n.
2 1x
n0
n0
n0
n0
The interval of convergence is x 2 < 1 or 1 < x < 1 since lim
n→ un1
2 x 2n2
lim
x2 .
n→ un
2 x2n
16. First finding the power series for 44 x, we have
1
1
x
1 14x n0 4
n
1n xn
4n
n0
Now replace x with x2.
1n x 2n
4
.
4 x2 n0
4n
The interval of convergence is x 2 < 4 or 2 < x < 2 since
lim
n→ 17.
un1
1 x
lim
n→ un
4n1
n1 2n2
4n
1n x 2n
x2
x2
.
4
4
1
1n x n
1 x n0
1
1n xn 12n x n xn
1 x n0
n0
n0
hx 1
1
2
1n x n xn 1n 1 x n
1 1 x 1 x n0
n0
n0
x2
2 0x 2x2 0x3 2x4 0x5 2x6 . . . 2x
n0
2n,
1 < x < 1 (See Exercise 15.)
323
324
Chapter 9
18. hx Infinite Series
1
1 n
1
x
1
1nxn x
2n0
x2 1 21 x 21 x 2n0
1
1
1n 1xn 0 2x 0x2 2x3 0x4 2x5 . . .
2n0
2
1
2x2n1 x2n1, 1 < x < 1
2n0
n0
19. By taking the first derivative, we have
d
1
x 12 dx
d
1
1
. Therefore,
dx x 1
x 12
1 x 1 nx
n
n
n0
n
n1
n1
1
n1
n 1 x n, 1 < x < 1.
n0
d2
2
x 13 dx 2
d
dx
1 x n n
n0
1 nx 1 nn 1x
n
n1
n1
21. By integrating, we have
lnx 1 d2
1
2
. Therefore,
dx 2 x 1
x 13
20. By taking the second derivative, we have
n
n2
n2
1 n 2n 1 x , 1 < x < 1.
n
n
n0
1
dx lnx 1. Therefore,
x1
1n x n dx C n0
1n x n1
, 1 < x ≤ 1.
n1
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
lnx 1 1n x n1
, 1 < x ≤ 1.
n1
n0
22. By integrating, we have
1
dx ln1 x C1 and
1x
1
dx ln1 x C2.
1x
f x ln1 x 2 ln1 x ln1 x. Therefore,
ln1 x 2 1
dx 1x
1
dx
1x
1n x n dx n0
C
n0
x n dx C1 x2n2
n 1, 1 < x < 1.
n0
23.
x2
1
1n x 2n 1n x 2n, 1 < x < 1
1 n0
n0
2x 2n2
1x 2n2
1n 1 xn1
C
C
n1
n1
n0
n0 2n 2
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
ln1 x2 x n1
1n x n1
C2 n
1
n0
n0 n 1
Section 9.9
24.
Representation of Functions by Power Series
325
2x
2x 1n x 2n (See Exercise 23.)
x2 1
n0
1 2x
n
2n1
n0
Since
2x
d
lnx 2 1 2
, we have
dx
x 1
lnx 2 1 1n 2x 2n1 dx C n0
1n x 2n2
, 1 ≤ x ≤ 1.
n1
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
lnx 2 1 1
1
1
1
1n x n, we have 2
1n 4x 2n 1n 4n x 2n 1n 2x2n, < x < .
x 1 n0
4x 1 n0
2
2
n0
n0
25. Since,
26. Since
1n x 2n2
, 1 ≤ x ≤ 1.
n1
n0
1
1
dx arctan2x, we can use the result of Exercise 25 to obtain
4x 2 1
2
arctan2x 2
1
dx 2
4x 2 1
1n 4n x 2n1 1
1
, < x ≤ .
2n
1
2
2
n0
1n 4n x 2n dx C 2
n0
To solve for C, let x 0 and conclude that C 0. Therefore,
1n 4nx 2n1 1
1
, < x ≤ .
2n 1
2
2
n0
arctan2x 2
27. x x2
x 2 x3
≤ lnx 1 ≤ x 2
2
3
x
x
5
S3
f
2
lnx 1
−4
8
x
S2
−3
28. x x2
x 2 x3
2
3
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
x 2 x3 x 4
≤ lnx 1
2
3
4
5
S5
f
x 2 x3 x 4 x5
≤ x 2
3
4
5
−4
8
S4
−3
x
x
x 2 x3 x 4
2
3
4
lnx 1
x
x2
2
x3
3
x4
4
x5
5
0.0
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
326
29.
Chapter 9
Infinite Series
1n1x 1n x 1 x 12 x 13 . . .
n
1
2
3
n1
(a)
(c) x 0.5:
3
n=3
12n
1n112n
0.693147
n
n
n1
n1
n=1
0
4
n=2
(d) This is an approximation of ln12 . The error is approximately 0. The error is less than the first omitted term,
151 251 8.7 1018.
n=6
−3
(b) From Example 4,
1nx 1n1
1n1x 1n
n
n1
n1
n0
ln x,
30.
0 < x ≤ 2, R 1.
1nx2n1
x3
x5
x . . .
3! 5!
n0 2n 1!
(a)
1n122n1
0.4794255386
2n 1!
n0
(c)
4
(d) This is an approximation of sin2 .
1
−6
6
The error is approximately 0.
−4
(b)
1nx2n1
sin x, R n0 2n 1!
31. gx x line
32. gx x Matches (c)
33. gx x Matches (d)
x3 x5
3
5
34. gx x Matches (a)
x3 x5 x7
,
3
5
7
Matches (b)
1
In Exercises 35-38, arctan x n
n0
35. arctan
x3
, cubic with 3 zeros.
3
x2n1
.
2n 1
1
1
142n1
1n
1
1
1n
. . .
2n1
4 n0
2n 1
4 192 5120
n0 2n 14
1
1
1
1
Since 5120 < 0.001, we can approximate the series by its first two terms: arctan 4 4 192 0.245.
arctan x 2 36.
1
n
x 4n2
2n 1
n
x 4n3
C, C 0
4n 32n 1
n0
arctan x 2 dx 1
n0
34
arctan x 2 dx 0
1
n0
n
344n3
34n3
1n
4n 32n 1 n0
4n 32n 144n3
27
2187
177,147
192 344,064 230,686,720
Since 177,147230,686,720 < 0.001, we can approximate the series by its first two terms: 0.134.
Section 9.9
12
0
1
arctan x 2
1
1
dx . . .
1n
x
4n 22n 124n2 8 1152
n0
1
< 0.001, we can approximate the series by its first term:
1152
x 2 arctan x 38.
x 2 arctan x dx n
x 2n4
2n 42n 1
1
1
1
. . .
2n 42n 122n4 64 1152
n
1
< 0.001, we can approximate the series by its first term:
1152
In Exercises 39– 43, use
41.
1
n0
Since
x nx , x < 1
n
n0
n1
n1
x
x nx n1 nx n, x < 1
2
1 x
n1
n1
42.
x1 x
x 2n 1x n 2n 1x n1, x < 1
2
1 x
n0
n0
(See Exercise 41.)
x < 1
x n,
n1
x 2 arctan x dx 0.016.
0
40.
n1
1x
1
x
1 x2 1 x2 1 x2
12
nx
1
x n, x < 1.
1 x n0
d
d
1
1
1 x2 dx 1 x
dx
0
arctan x 2
dx 0.125.
x
x 2n3
2n 1
1
0
12
n
n0
x 2 arctan x dx 1
n0
12
39.
arctan x 2
x 4n2
1n
C Note: C 0
dx x
4n 22n 1
n0
Since
2n 1x , x < 1
n
n0
43. Pn En 12
n
n1
n1
nPn n2
1
n
1 1
n
2 n1 2
n1
1
1
2
2 1 122
1
Since the probability of obtaining a head on a single toss is 2 , it is expected that, on average, a head will be obtained
in two tosses.
In Exercise 44,
44. (a)
(b)
327
arctan x 2 1 x22n1
x 4n1
1n
12
x
x n0
2n 1
2n
1
n0
37.
Representation of Functions by Power Series
1
x n, x 1.
1 x n0
1 2
n
3 n1 3
n
1 9
n
10 n1 10
n
2 2
n
9 n1 3
n1
9 9
n
100 n1 10
2
1
2
9 1 232
n1
9
100
1
1 9102 9
328
Chapter 9
Infinite Series
45. Replace x with x.
47. Replace x with x and
multiply the series by 5.
46. Replace x with x2.
48. Integrate the series and
multiply by 1.
49. Let arctan x arctan y . Then,
tanarctan x arctan y tan tanarctan x tanarctan y
tan 1 tanarctan x tanarctan y
xy
tan 1 xy
arctan
1x xyy . Therefore, arctan x arctan y arctan1x xyy for xy 1.
50. (a) From Exercise 49, we have
1
120
1
120
arctan
arctan
arctan 119
239
119
239
arctan
arctan
(b) 2 arctan
1239
28,561
arctan arctan 1 1120119
1201191239
28,561 4
10
1
1
215
5
1
arctan arctan arctan
arctan
arctan
5
5
5
24
12
1 152
4 arctan
120
1
1
5
5
2512
1
2 arctan 2 arctan arctan
arctan
arctan
arctan
5
5
5
12
12
119
1 5122
4 arctan
1
1
120
1
arctan
arctan
arctan
(see part (a).)
5
239
119
239
4
51. (a) 2 arctan
2 arctan
1
1
1
1
1 0.53 0.55 0.57
1 173 175 177
4 arctan 8 4 3.14
2
7
2
3
5
7
7
3
5
7
1
1
arctan
2
3
(b) 4 arctan
2 4
1
lnx 1 1
123 125 127
1 133 135 137
4 40.4635 40.3217 3.14
3
5
7
3
3
5
7
1n1xn
1n xn1
n1
n
n0
n1
1n1xn
.
n
n1
n1
1n112n
1
2n n n1
n
ln
54. From Exercise 53, we have
53. From Exercise 21, we have
n1
1
1
12 13
56
arctan arctan
arctan
2
3
1 1213
56
4
52. (a) arctan
1
4
1
43 17
1
25
arctan arctan arctan arctan
arctan 1 arctan
2
7
3
7
1 4317
25
4
(b) 8 arctan
Thus,
1
1
4
1
2 2
arctan arctan arctan
arctan
2
2
2
1 122
3
12 1 ln 23 0.4055.
1
n1
n1
1n113n
1
n
3 n n1
n
ln
13 1 ln 34 0.2877.
Section 9.10
1n1
n1
1
n
n0
n0
2n1
2
x 2n1
.
2n 1
2n1
0.7854
4
58. From Exercise 56, we have
1
12
1n
2n 1 n0
2n 1
n
329
1
1
1n
2n 1 n0
2n 1
arctan 1 57. From Exercise 56, we have
1
n
n0
1n125n
2n
n 5 n n1
n
2
7
ln 1 ln 0.3365.
5
5
1
56. From Example 5, we have arctan x 55. From Exercise 53, we have
Taylor and Maclaurin Series
arctan
2n1
1
n1
n1
1
1
1n 2n1
2n 1 n0
3
2n 1
3
2n1
1
0.4636.
2
1
n
n0
arctan
132n1
2n 1
1
0.3218.
3
59. f x arctan x is an odd function (symmetric to the origin).
60. The approximations of degree 3, 7, 11, . . . ,
4n 1, n 1, 2, . . . have relative extrema.
61. The series in Exercise 56 converges to its sum at a slower
rate because its terms approach 0 at a much slower rate.
62. Because
63. Because the first series is the derivative of the second
series, the second series converges for x 1 < 4 and
perhaps at the endpoints, x 3 and x 5.
64. Using a graphing utility, you obtain the following partial
sums for the left hand side. Note that 1 0.3183098862.
d a xn nan x n1, the radius of
dx n0 n
n1
convergence is the same, 3.
n 0: S0 0.3183098784
n 1: S1 0.3183098862
65.
1n
n2n 1
3
n0
66.
From Example 5 we have arctan x 1
n0
1n
3 2n 1 n0
n
n0
n
x2n1
.
2n 1
3
1n
32n2n 1 3
1n1
32n1
2n 1
n0
3
3 arctan
3
Section 9.10
13
6 0.9068997
Taylor and Maclaurin Series
1. For c 0, we have:
f x e2x
f nx 2n e2x ⇒ f n0 2n
e2x 1 2x 2xn
4x 2 8x3 16x 4 . . .
.
2!
3!
4!
n0 n!
1n 2n1
32n1
1n
2n 1! n0
2n 1!
n0 3
2n1
sin
3 3
2
0.866025
330
Chapter 9
Infinite Series
2. For c 0, we have:
f x e3x
f nx 3ne 3x ⇒ f n0 3n
e 3x 1 3x 3xn
9x2 27x3 . . .
.
2!
3!
n0 n!
3. For c 4, we have:
f x cosx
f
4 fx sinx
f
4 22
f x cosx
f
4 22
fx sinx
f
f 4x cosx
f 4
2
2
4 2
4 2
2
2
and so on. Therefore, we have:
cos x f n4x 4n
n!
n0
2
2
1 x 4 x 2!4
2
x 43 x 44 . . .
3!
4!
1nn12x 4n
.
2 n0
n!
2 [Note: 1nn12 1, 1, 1, 1, 1, 1, 1, 1, . . .]
4. For c 4, we have:
f x sin x
f
4 2
fx cos x
f
4 2
f x sin x
f
4 22
fx cos x
f
4 22
f 4
4 f 4x sin x
2
2
2
2
and so on. Therefore we have:
sin x f n4x 4n
n!
n0
2
2
2
2
1 x 4 x 2!4
2
1
n0
x 43 x 44 . . .
3!
4!
x 4n1
1 .
n 1!
nn12
Section 9.10
Taylor and Maclaurin Series
5. For c 1, we have,
f x ln x
fx 1
x
f1 1
1
x2
f x f 1 1
2
x3
fx f 4x f 5x f 1 0
f1 2
6
x4
f 41 6
24
x5
f 51 24
and so on. Therefore, we have:
ln x f n1x 1n
n!
n0
0 x 1 x 1 1
n
n0
x 12 2x 13 6x 14 24x 15 . . .
2!
3!
4!
5!
x 12 x 13 x 14 x 15 . . .
2
3
4
5
x 1n1
.
n1
6. For c 1, we have:
f x ex
f nx ex ⇒ f n1 e
ex x 1n
f n1x 1n
x 12 x 13 x 14 . . .
e 1 x 1 e
.
n!
2!
3!
4!
n!
n0
n0
7. For c 0, we have:
f x sin 2x
f 0 0
fx 2 cos 2x
f0 2
f x 4 sin 2x
f 0 0
fx 8 cos 2x
f0 8
f 4x 16 sin 2x
f 40 0
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
and so on. Therefore, we have:
sin 2x f n0xn
0x 2 8x3 0x 4 32x5 0x6 128x7 . . .
0 2x n!
2!
3!
4!
5!
6!
7!
n0
2x 1n2x2n1
8x3 32x5 128x7 . . .
.
3!
5!
7!
2n 1!
n0
331
332
Chapter 9
Infinite Series
8. For c 0, we have:
f x lnx 2 1
f 0 0
fx 2x
x2 1
f0 0
f x 2 2x 2
x2 12
f 0 2
f x 4xx 2 3
x2 13
f 0 0
f 4x 12x 4 6x 2 1
x 2 14
f 40 12
f 5x 48xx 4 10x 2 5
x 2 15
f 50 0
f 6x 2405x6 15x 4 15x 2 1
x 2 16
f 60 240
and so on. Therefore, we have:
lnx 2 1 f n0xn
2x 2 0x3 12x 4 0x5 240x6 . . .
0 0x n!
2!
3!
4!
5!
6!
n0
x2 1n x 2n2
x 4 x6 . . .
.
2
3
n1
n0
9. For c 0, we have:
f x secx
f 0 1
fx secx tanx
f0 0
f x sec3x secx tan2x
f 0 1
f x 5 sec3x tanx secx tan3x
f 4x
5
secx x 18
sec5
sec3
x
x secx
tan2
f 0 0
tan4
x
f 40 5
f n0xn
x 2 5x 4 . . .
1 .
n!
2!
4!
n0
10. For c 0, we have;
f x tanx
f 0 0
fx sec2x
f0 1
f x 2 sec2x tanx
f 0 0
f x 2sec4x 2 sec2x tan2x
f 0 2
f 4x 8sec4x tanx sec2x tan3x
f 40 0
f 5x 82 sec6x 11 sec4x tan2x 2 sec2x tan4x
f 50 16
tanx f n0xn
2x3 16x5 . . .
x3
2
x
x x5 . . ..
n!
3!
5!
3
15
n0
Section 9.10
Taylor and Maclaurin Series
333
1n x2n
.
n0 2n!
11. The Maclaurin series for f x cos x is
Because f n1x ± sin x or ± cos x, we have f n1z ≤ 1 for all z. Hence by Taylor’s Theorem,
0 ≤ Rn x Since lim
n→ f n1z
n 1!
xn1 ≤
x .
n 1!
n1
xn1 0, it follows that Rnx → 0 as n → . Hence, the Maclaurin series for cos x converges to cos x for all x.
n 1!
12. The Maclaurin series for f x e2x is
2xn
.
n!
n0
f n1x 2n1e2x. Hence, by Taylor’s Theorem,
0 ≤ Rnx Since lim
n→ f n1z n1
2n1e2z n1
x
x
.
n 1!
n 1!
2n1xn1
2xn1
lim
0, it follows that Rnx → 0 as n → .
n→ n 1!
n 1!
Hence, the Maclaurin Series for e2x converges to e2x for all x.
13. The Maclaurin series for f x sinh x is
x 2n1
2n 1!.
n0
f
n1x
sinh x or cosh x. For fixed x,
0 ≤ Rnx f n1z n1
sinhz n1
x
x
→ 0 as n → .
n 1!
n 1!
The argument is the same if f n1x cosh x. Hence, the Maclaurin series for sinh x converges to sinh x for all x.
14. The Maclaurin series for f x cosh x is
x 2n
2n!.
n0
f n1x sinh x or cosh x. For fixed x,
0 ≤ Rnx f n1z n1
sinhz n1
x
x
→ 0 as n → .
n 1!
n 1!
The argument is the same if f
15. Since 1 xk 1 kx 1 x2 1 2x n1x
cosh x. Hence, the Maclaurin series for cosh x converges to cosh x for all x.
kk 1x 2 kk 1k 2x3 . . .
, we have
2!
3!
23x 2 234x3 2345x 4 . . .
1 2x 3x 2 4x3 5x 4 . . .
2!
3!
4!
1 n 1x .
n0
kk 1x2 kk 1k 2x3 . . .
, we have
2!
3!
2
1
1232x
123252x3 . . .
1
x
2
2!
3!
16. Since 1 xk 1 kx 12
1 x
1
x
13x 2 135x3 . . .
2
222!
233!
1
n1
1
35.
. . 2n 1xn
2n n!
.
n
n
334
17.
Chapter 9
Infinite Series
2 12
1 2x 1
1
2
2
4 x
and since 1 x12 1 1n 1
1
1
1
2
2
4 x
n1
1n 1 3 5 . . . 2n 1xn
, we have
2n n!
n1
35.
. . 2n 1x22n
1n 1
1
2 n!
2 n1
n
35.
. . 2n 1x2n
.
n!
3n1
2
1
1434 2 143474 3 . . .
18. 1 x14 1 x x x 4
2!
3!
1
37
3 7 11 4 . . .
3
1 x 2 x2 3 x3 x 4
4 2!
4 3!
444!
1n1 3
1
1 x
4
n2
19. Since 1 x12 1 1n1 1
x
2 n2
we have 1 x 212 1 20. Since 1 x12 1 ex xn
35.
35.
x4
. . 2n 3x 2n
2n n!
35.
.
. . 2n 3xn
2n n!
x3
. . 2n 3xn
2 n!
1n1 1
x3
2
n2
x2
xn
n
. . 4n 5
4nn!
1n1 1
x2
2
n2
1n1 1
x
2 n2
we have 1 x312 1 21.
7 11 .
35.
. . 2n 3x3n
2n n!
.
x5
n! 1 x 2! 3! 4! 5! . . .
n0
ex 2 2
22.
ex x 2n
x 22n
x2
x4
x6
x8
1
2 3 4 . . .
n
n!
2
2 2! 2 3! 2 4!
n0
n0 2 n!
xn
x2
x3
x4
x5
n! 1 x 2! 3! 4! 5! . . .
n0
e3x 23.
1n 3n xn
3xn
9x 2 27x3 81x 4 243x5 . . .
1 3x n!
n!
2!
3!
4!
5!
n0
n0
sin x sin 3x 1nx2n1
n0 2n 1!
24.
1n3x2n1
2n 1!
n0
cos x cos 4x 1nx2n
x2
x4
x6
1 . . .
2! 4! 6!
n0 2n!
1n42nx2n
1n4x2n
2n!
2n!
n0
n0
1
25.
cos x cos x32 1n x 2n
x2 x 4 . . .
1 2n
!
2!
4!
n0
1n x322n
2n!
n0
1n x3n
2n!
n0
1
x3
x6
. . .
2! 4!
26.
sin x 16x2 256x4 . . .
2!
4!
1nx2n1
n0 2n 1!
1nx32n1
n0 2n 1!
2 sin x3 2
x9
x15 . . .
3!
5!
2x3 2x9 2x15 . . .
3!
5!
2 x3 Section 9.10
27.
ex 1 x x2
x3
x4
x5
. . .
2! 3!
4!
5!
ex 1 x x2
x3
x4
x5
. . .
2! 3!
4!
5!
e x ex 2x 28.
2x3 2x5 2x7 . . .
3!
5!
7!
ex 1 x x3
x2
. . .
2! 3!
ex 1 x x2
x3
. . .
2! 3!
2x2 2x4 . . .
2!
4!
ex ex 2 1 x
e ex
2
sinhx Taylor and Maclaurin Series
2 cos hx ex ex x2n
2 2n!
n0
x2n1
x5
x7
x3
x . . .
3! 5! 7!
n0 2n 1!
1
29. cos2x 1 cos2x
2
1n2x2n
1
1
2x2 2x4 2x6 . . .
1
11
2
2!
4!
6!
2
2n!
n0
30. The formula for the binomial series gives 1 x12 1 1 x 212 1 ln x x 2 1 31. x sin x x x x2 33.
1n 1 3 5 . . . 2n 1x 2n
2n n!
n1
dx
1n1 3 5 . . . 2n 1x2n1
2n2n 1n!
n1
x3
2
3
x3
x5
. . .
3! 5!
1 3x5
1 3 5x7
. . ..
245 2467
32. x cos x x 1 x4
x6
. . .
3! 5!
x
1nx2n2
n0 2n 1!
sin x x x33! x55! . . .
x
x
1
35.
1n 1 3 5 . . . 2n 1xn
, which implies that
2n n!
n1
1
x
34.
1nx2n
x2
x4
. . .
,x0
2! 4!
n0 2n 1!
eix 1 ix eix 1 ix eix eix 2ix x 2 1
x
x2
x4
. . .
2! 4!
x3
x5
. . .
2! 4!
1nx2n1
2n!
n0
arcsin x
2n!x2n1
n
x
2
n!22n 1
n0
1
x
2n!x2n
2 n! 2n 1, x 0
n0
n
2
x 2 ix3 x 4 ix5
x6
ix2 ix3 ix4 . . .
1 ix . . .
2!
3!
4!
2!
3!
4!
5!
6!
x 2 ix3 x 4 ix5
x6
ix2 ix3 ix4 . . .
1 ix . . .
2!
3!
4!
2!
3!
4!
5!
6!
2ix3 2ix5 2ix7 . . .
3!
5!
7!
1n x2n1
eix eix
x3
x5
x7
x . . .
sinx
2i
3! 5! 7!
n0 2n 1!
335
336
Chapter 9
Infinite Series
2x 2 2x 4 2x6 . . .
2!
4!
6!
36. eix eix 2 (See Exercise 35.)
1n x 2n
eix eix
x2 x 4
x6
1 . . .
cosx
2
2!
4!
6!
2n!
n0
37. f x ex sin x
14
1x
x 2 x3
x4
. . .
2
6
24
x x2 x x2 x3
x5
. . .
3
30
x
. . .
x x6 120
3
5
P5
x3 x3
x4 x4
x5
x5
x5
. . .
2
6
6
6
120 12 24
f
−6
6
−2
38. gx ex cos x
8
x2 x 4
x4
. . .
2
6
24
1x
1 x2 24x . . .
2
−6
4
6
g
P4
x4
x4
x2 x2
x3 x3
x4
1x
. . .
2
2
6
2
24
4
24
1x
x3
3
x4
6
−40
. . .
39. hx cos x ln1 x
1
4
x2
x4
. . .
2
24
x
x 2 x3 x 4 x5 . . .
2
3
4
5
P5
−3
9
h
x
x2
x3 x3
x4 x4
x5 x5
x5
. . .
2
3
2
4
4
5
6
24
x
x 2 x3 3x5 . . .
2
6
40
−4
40. f x ex ln1 x
1x
x4
x 2 x3
. . .
2
6
24
x x2 x
3
−3
P5
3
−3
2
3
4
5
x2
x3 x3 x3
x4 x4 x4 x4
x5 x5 x5
x5
x5
. . .
2
3
2
2
4
3
4
6
5
4
6
12 24
x 2 x3 3x5 . . .
2
3
40
f
x x2 x3 x4 x5 . . .
Section 9.10
41. gx sin x
. Divide the series for sin x by 1 x.
1x
Taylor and Maclaurin Series
gx x x2 5x3 5x4
6
6
x3
x5
0x4 . . .
1 x x 0x2 6
120
x x2
x3
x2 6
x2 x3
5x3
0x4
6
5x4
5x3
6
6
x5
5x4
6
120
5x4 5x5
6
6
5x 3 5x 4 . . .
6
6
4
x x2 g
−6
6
P4
−4
42. f x ex
. Divide the series for ex by 1 x.
1x
x2 x3
2
3
x2
1x 1x 2
1x
x2
0
2
x2
2
1
f x 1 3x 4 . . .
8
x3
x4
x5
. . .
6
24
120
x 2 x 3 3x 4 . . .
2
3
8
6
P4
−6
f
6
−6
x3
6
x3
2
x3
x4
3
24
x3
x4
3
3
3x 4
x5
8
120
3x4 3x5
8
8
43.
45.
y x2 x4
x3
x x
3!
3!
44.
yx
f x x sin x
f x x cos x
Matches (c)
Matches (d)
y x x2 f x xe x
Matches (a)
x3
x2
x 1x
2!
2!
46.
x3
x5
x2 x 4
x 1 2! 4!
2!
4!
y x 2 x3 x 4 x 21 x x 2
f x x 2
1
1x
Matches (b)
337
338
Chapter 9
Infinite Series
x
47.
0
0
1n1t2n2
dt n 1!
n0
x
0
x
48.
x
1 t3 dt 0
1
0
1n11
t3
2
n2
1n1t2n3
2n 3n 1!
x
35.
0
n0
. . 2n 3t3n
2n n!
1n11 3 5 . . . 2n 3t3n1
t4
8
3n 12n n!
n2
x
1n11 3 5 . . . 2n 3x3n1
x4
8
3n 12n n!
n2
1n1x 2n3
2n 3n 1!
n0
dt
x
0
1nx 1n1
x 12 x 13 x 14 . . .
x 1 ,
n1
2
3
4
n0
we have ln 2 1 1 1 1 . . .
1
1n1 0.6931. (10,001 terms)
2 3 4
n
n1
1nx 2n1
x3
x5
x7
x . . . , we have
2n
1
!
3!
5!
7!
n0
1n
0 < x ≤ 2
50. Since sinx sin1 t
49. Since ln x 1n t2n
1 dt
n!
n0
x
et 2 1 dt 1
1
x2
x3
1
2n 1! 1 3! 5! 7! . . . 0.8415.
(4 terms)
n0
51. Since ex xn
n! 1 x 2! 3! . . . ,
n0
we have e2 1 2 52. Since ex xn
2n
22 23 . . .
7.3891. (12 terms)
2! 3!
n0 n!
x2
x3
x4
x5
n! 1 x 2! 3! 4! 5! . . . , we have e
1
11
n0
and
1
1
1
1
. . .
2! 3! 4! 5!
1n1
e1
1
1
1
1
1
1 e1 1 . . . 0.6321. (6 terms)
e
2! 3! 4! 5! 7!
n!
n1
54. Since
53. Since
cos x 1nx 2n
x 2 x 4 x6 x8
1 . . .
2n
!
2! 4! 6! 8!
n0
1 cos x 1nx 2n2
x2 x 4
x6
x8
. . .
2!
4!
6! 8!
n0 2n 2!
1nx 2n1
1 cos
x
x3
x5
x7
. . .
x
2! 4! 6! 8!
n0 2n 2!
we have lim
x→0
1x 2n1
1 cos x
lim
0.
x→0 n0 2n 2!
x
sin x 1n x 2n1
x3
x5
x7
x . . .
2n
1
!
3!
5!
7!
n0
1nx 2n
sin x
x2 x 4
x6
1 . . .
x
3!
5!
7!
n0 2n 1!
we have lim
x→0
1n x 2n
sin x
lim
1.
x→0 n0 2n 1!
x
Section 9.10
1
55.
0
sin x
dx x
Since 17
1
1nx 2n
dx n0 2n 1!
1
0
7!
sin x
1.
x
1 x3 x5 x7 . . . dx x 3x
arctan x
dx x
0
Since 1
0
12
2
4
6
3
2
0
x5
x7
. . .
52 72
12
0
< 0.0001, we have
arctan x
1
1
1
1
1
0.4872.
dx x
2 3223 5225 7227 9229
Note: We are using lim
x→0
arctan x
1.
x
x
5x
. . .
1 x2 x8 16x 5x128 . . . dx x x8 56x 160
1664
0.3
0.3
1 x3 dx 0.1
0.1
1
7
56 0.3
Since
0
0
145
< 0.0001,
Since
15
x cos x dx 0
0
1
1
cos x dx 0.5
10
0.3
13
0.1
0.14 0.201.
x3 x4 x5 . . .
x3
x4
x5
x6
dx . . .
2
3
4
3
42 53 64
1nx4n12
dx 2n!
n0
10!
x cos x dx 2
0.5
7
14
0
143 144
0.00472.
x lnx 1 dx 3
8
0
2
1
4
14
Since 2223223
x
2
0
12
4
14
x lnx 1 dx 9
0.3 0.1 81 0.3
1 x3 d x 1
4
2
6
0.1 < 0.0001, we need two terms.
0.1
3
7
0.3
60.
1n
n0 2n 12n 1!
9229
12
59.
0
Note: We are using lim
12
58.
< 0.0001, we need three terms:
x→0
57.
1
sin x
1
1
dx 1 . . . 0.9461. using three nonzero terms
x
3 3! 5 5!
0
56.
1nx 2n1
n0 2n 12n 1!
Taylor and Maclaurin Series
1nx4n32
4n 3
2n!
2
n0
2
0
1n2x4n32
n0 4n 32n!
2
0
< 0.0001, we need five terms.
232 272 2112 2152 2192
0.7040.
3
14
264
10,800
766,080
1 2!x 4!x 6!x 8!x . . . dx x 2x2! 3x4! 4x6! 5x8! . . .
2
3
4
2
3
4
5
1
0.5
1
Since 201,600 1 0.55 < 0.0001, we have
1
cos x dx 1 0.5 0.5
61. From Exercise 21, we have
1
2
1
0
2
ex 2 dx 1
2
1
1
1
1
1 0.52 1 0.53 1 0.54 1 0.55 0.3243.
4
72
2880
201,600
1
0
1nx 2n
1
dx nn!
2
2
n0
1
2
x
21
n!2n 1
n0
n 2n1
n
1
0
1 2 1 3 2
1
1
2
2! 5
1n
1
2
2 n!2n 1
n0
n
3! 7
1
23
0.3414.
339
340
Chapter 9
Infinite Series
62. From Exercise 21, we have
1
2
2
ex 2dx 2
1
1
2
2
1
f x x cos 2x P5x x 2x3 1n x 2n1
2 n!2n 1
n0
2
n
1
1 1n2n1 1
2 n0 2nn!2n 1
1 2 1 3 2
7
1
2
31
2
63.
1
1nx 2n
dx 2nn!
2
n0
8191
26
6! 13
2! 5
127
23
3! 7
511
24
4! 9
2047
25
5! 11
32,767
131,071
524,287
0.1359.
27 7! 15 28 8! 17 29 9! 19
1n4nx2n1
2n!
n0
64. f x sin
2x5
3
x
ln1 x
2
x 2 x3 7x 4 11x5
2
4
48
96
P5x 4
2
f
f
−3
−4
3
8
P5
P5
−2
−4
The polynomial is a reasonable approximation on the
interval 0.60, 0.73.
The polynomial is a reasonable approximation on the
interval 34 , 34.
65.
f x x ln x, c 1
P5x x 1 3
g
P5
x 13 x 14 71x 15
24
24
1920
−2
4
The polynomial is a reasonable approximation on the interval 4 , 2.
1
66.
−2
3
f x x arctan x, c 1
3
P5x 0.7854 0.7618x 1 0.3412
x 2! 1 0.0424 x 3! 1 2
x 14
x 15
1.3025
5.5913
4!
5!
3
f
P5
−2
4
−1
The polynomial is a reasonable approximation on the interval 0.48, 1.75.
67. See Guidelines, page 680.
68. a2n1 0 (odd coefficients are zero)
69. (a) Replace x with x.
(c) Multiply series by x.
(b) Replace x with 3x.
70. The binomial series is 1 xk 1 kx (d) Replace x with 2x, then replace x with 2x, and add the
two together.
kk 1 2 kk 1k 2 3 . . .
x x . The radius of convergence is R 1.
2!
3!
Section 9.10
g
g
kx
x 2 ln 1 kv0 cos k
v0 cos 71. y tan Taylor and Maclaurin Series
2
3
tan x gx
gx
gx 2
gkx3
gk2x 4
. . .
2
3
2
3
kv0 cos kv0 cos 2v0 cos 3v0 cos 4v04 cos4 tan x gx2
kgx3
k2gx 4
. . .
2v02 cos2 3v03 cos3 4v04 cos4 . . .
1
, g 32
16
32x 2
11632x3 116232x 4 . . .
2
2
264 12
3643123
4644124
2264x 2 2
3x 32
3x 32
2
24x 4
23x3
. . .
3
364 16 4644162
2n xn
n64 16
n
n2
73. f x 4
72. 60, v0 64, k y 3x kx
1
4 v0 cos kx
gx
g
kx
1
kv0 cos k2 v0 cos 2 v0 cos kx
1
3 v0 cos tan x e1x ,
0,
2
n2
3x 32
xn
n32 16
n
n2
n2
x0
x0
f x f 0
e1x 0
lim
x→0
x0
x
2
(a)
(b) f0 lim
y
x→0
2
e1x
. Then
x→0
x
2
1
Let y lim
x
−3 − 2 −1
1
2
3
ln y lim ln
x→0
e1x
x
2
lim x1 ln x lim 1 x x
x→0
2
2
ln x
2
x→0
Thus, y e 0 and we have f0 0.
(c)
f n0 n
f0x f 0x 2 . . .
x f 0 0 f x This series converges to f at x 0 only.
n!
1!
2!
n0
74. (a) f x lnx 2 1
.
x2
(b)
1.5
From Exercise 8, you obtain:
P
1nx 2n
1 1nx 2n2
x 2n0 n 1
n0 n 1
2
4
lnt2 1
dt
t2
x
0
x
Gx 6
2
−0.5
8
x
x
x
x
.
2
3
4
5
P8 1 (c) Fx 0
P8t d t
x
0.25
0.50
0.75
1.00
1.50
2.00
Fx
0.2475
0.4810
0.6920
0.8776
1.1798
1.4096
Gx
0.2475
0.4810
0.6924
0.8865
1.6878
9.6063
0
(d) The curves are nearly identical for 0 < x < 1. Hence, the integrals nearly agree on that interval.
.
341
342
Chapter 9
Infinite Series
75. By the Ratio Test: lim
n →
76. ln
xn1
n 1!
n!
xn
lim
n →
x
n1
0 which shows that
xn
n! converges for all x.
n0
11 xx ln1 x ln1 x
x
x2 x3 . . .
2
3
2x 2
ln 3 ln
x x2 x3 . . . 2
3
x3
x5
x 2n
2 . . . 2x
,R1
3
5
2n
1
n0
1
12
2 1 11 12
12 2
3
2
1
1
124 126
1
1.098065
1
5
7
12 80 448
ln 3 1.098612
77.
53 5 3!4 3 606 10
78.
22 22!3 3
79.
0.54 0.50.54!1.52.5
0.0390625 80.
134373103133
13
5 5!
81. 1 xk k
n
n0
91
0.12483
729
nx
Example: 1 x2 nx
2
n
1 2x x 2
n0
82. Assume e pq is rational. Let N > q and form the following.
e 11
1
1
1
1
. . .
. . .
2!
N!
N 1! N 2!
Set a N! e 1 1 . . . 1
N!
, a positive integer. But,
N 1 1! N 1 2! . . . N 1 1 N 11 N 2 . . . < N 1 1 N 1 1
a N!
83. gx 2
1
1
1
1
1
. . . N1
N 1 N 12
N1
1
1
1
N1
x
a0 a1x a2 x 2 . . .
1 x x2
x 1 x x2a0 a1x a2 x 2 . . .
x a0 a1 a0x a2 a1 a0x2 a3 a2 a1x3 . . .
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 an1 an2. The coefficients are the Fibonacci numbers.
1
, a contradiction.
N
. . .
5
128
Review Exercises for Chapter 9
343
84. Assume the interval is 1, 1. Let x 1, 1,
f 1 f x 1 x f(x) 121 x2 f (c), c x, 1
f 1 f x 1 x f(x) 121 x2 f (d), d 1, x.
Hence, f 1 f 1 2 fx 121 x2 f c 121 x2 f d
2 fx f 1 f 1 121 x2f c 121 x2f d.
2 fx ≤ f 1 f 1 121 x2 f c 121 x2 f d
Since f x ≤ 1 and f x ≤ 1,
≤ 1 1 21 x 2 21 x2
1
1
3 x 2 ≤ 4.
Thus, f x ≤ 2.
Note: Let f x 12 x 12 1. Then f x ≤ 1, f x 1 and f 1 2.
Review Exercises for Chapter 9
1. an 1
n!
2. an 2
3. an 4 : 6, 5, 4.67, . . .
n
Matches (a)
5. an 100.3n1: 10, 3, . . .
n
4. an 4 : 3.5, 3, . . .
2
Matches (c)
7. an n
n2 1
Matches (d)
5n 2
n
The sequence seems to
converge to 5.
5n 2
lim a n→
lim
n→ n
n
n
2
The sequence seems to diverge
(oscillates).
n
sin
: 1, 0, 1, 0, 1, 0, . . .
2
8. an sin
0
n1
: 6, 4, . . .
Matches (b)
8
n→
lim 5 32
6. an 6 12
0
2
0
12
−2
2
5
n
n1
0
n→ n2
1
0
n→ n
Converges
10. lim
9. lim
Converges
n3
n→ n 1
11. lim
12. lim
n→ 2
n
1
lim
lnn n→ 1n
Diverges
13. lim n 1 n lim n 1 n n→ n→ 14. lim 1 n→ 1
2n
n
lim
k→ Converges; k 2n
1 k 1
k 12
e12
n 1 n
n 1 n
lim
n→ 1
n 1 n
sin n
0
n
Converges
15. lim
n→ 0
Converges
344
Chapter 9
Infinite Series
ln y 0.05
4
17. An 5000 1 y bn cn1n
16. Let
n 1, 2, 3
lnbn cn
n
n
50001.0125n
(a) A1 5062.50
A5 5320.41
A2 5125.78
A6 5386.92
A3 5189.85
A7 5454.25
A4 5254.73
A85522.43
1
bn ln b cn ln c.
lim ln y lim n
n→ n→ b cn
Assume b ≥ c and note that the terms
bn ln b cn ln c
bn ln b
cn ln c
n
n
n
n
n
b c
b c
b cn
(b) A40 8218.10
converge as n → . Hence an converges.
18. (a) Vn 120,0000.70n, n 1, 2, 3, 4, 5
(b) V5 120,0000.705 $20,168.40
19. (a)
(b)
k
5
10
15
20
25
Sk
13.2
113.3
873.8
6448.5
50,500.3
The series diverges geometric r 32 > 1.
20. (a)
120
0
12
− 10
(b)
k
5
10
15
20
25
Sk
0.3917
0.3228
0.3627
0.3344
0.3564
The series converges by the Alternating Series Test.
1
0
12
0
21. (a)
(b)
k
5
10
15
20
25
Sk
0.4597
0.4597
0.4597
0.4597
0.4597
0
The series converges by the Alternating Series Test.
22. (a)
12
− 0.1
(b)
k
5
10
15
20
25
Sk
0.8333
0.9091
0.9375
0.9524
0.9615
The series converges, by the Limit Comparison Test with
1
1
1
n .
0
2
12
0
23. Converges. Geometric series, r 0.82, r < 1.
24. Diverges. Geometric series, r 1.82 > 1.
25. Diverges. nth-Term Test. lim an 0.
2
26. Diverges. nth-Term Test, lim an .
n→ 3
n→ 27.
3
2
n
n0
S
2
Geometric series with a 1 and r .
3
a
1
1
3
1 r 1 23 13
28.
2
2n2
n 4
3
n0
n0 3
See Exercise 27.
n
43 12
Review Exercises for Chapter 9
29.
2
1
n
n0
30.
3
2
n
1
1
3n
n0 2
n0
n
3
1
n
n0
2
n 1n 2
3
1
1
1
3 1
2 1 12 1 13
2 2
n
n0
345
n 1 n 2
1
1
n0
1
1 23
1 21 12 31 13 41 . . . 3 1 2
31. (a) 0.09 0.09 0.0009 0.000009 . . . 0.091 0.01 0.0001 . . . 0.090.01
n
n0
(b) 0.09 0.09
1
1 0.01 11
32. (a) 0.923076 0.923076
1 0.000001 0.0000012 . . . 0.9230760.000001
n
n0
(b) 0.923076 0.923076
923,076 1276,923 12
1 0.000001 999,999 1376,923 13
33. D1 8
39
32,0001.055
34. S n
n0
D2 0.78 0.78 160.7
32,0001 1.05540
1 1.055
$4,371,379.65
D 8 160.7 160.72 . . . 160.7n . . .
8 160.7
n
8 n0
16
4513 meters
1 0.7
35. See Exercise 110 in Section 9.2.
A
36. See Exercise 110 in Section 9.2.
Pe rt 1
er12 1
AP
200e0.062 1
e0.0612 1
100
$5087.14
37.
x4 lnx dx lim
b→ ln3xx 9x1 b
3
3
0
1
1 1
9 9
38.
n
n1
1
2
1
1
1
2 n
n1 n
n1 n
n
n1
lim
n→ 3
1
2n
1n3 2n
n32
lim
1
32
n→
n3 2n
1n By a limit comparison test with the convergent p-series
1
the series converges.
32 ,
n1 n
1
n
n1
40.
4
3
n
n1
Since the second series is a divergent p-series while
the first series is a convergent p-series, the difference
diverges.
41.
1
12
0.035
1 0.035
12 120
1
n1
1
n34
3
Divergent p-series, p 4 < 1
By the Integral Test, the series converges.
39.
12t
$14,343.25
1
12r 1 12r 1
2
1
1
1
2
n
2n
n1 n
n1 2
The first series is a convergent p-series and the second
series is a convergent geometric series. Therefore, their
difference converges.
42.
n1
nn 2
n1
lim
n→ n 1nn 2
n1
lim
1
n→ n 2
1n
By a limit comparison test with
1
n, the series diverges.
n1
346
43.
Chapter 9
3 5 . . . 2n 1
2 4 6 . . . 2n
1
n1
1
an Infinite Series
1
3
44. Since
n1
n
converges,
3
n1
n
1
converges by the
5
Limit Comparison Test.
3 5 . . . 2n 1
2 4 6 . . . 2n
1
1 1
>
32 54 . . . 2n
2n 2 2n 2n
1
1
1
2n 2 n diverges (harmonic series),
Since
n1
n1
so does the original series.
45. Converges by the Alternating Series Test.
(Conditional convergence)
46.
1nn
n1 n 1
an1 n 1
n2
≤
n
an
n1
n
0
n1
lim
n→
By the Alternating Series Test, the series converges.
48. Converges by the Alternating Series Test.
47. Diverges by the nth-Term Test.
an1 49.
n
e
50.
n2
n1
lim
n→ an1
n1
lim n12
n→ e
an
lim
n→ lim
n→ en2
n
e n 1
en2 2n1n
n2
n!
e
n1
3 lnn 1 3 ln n
3 ln n
<
an, lim
0
n→ n1
n
n
lim
n→ n
an1
n 1!
lim
n→ an
en1
lim
n→ e 1 n n 1
n1
e
By the Ratio Test, the series converges.
2n
n
3
n1
lim
n→ an1
2n1
lim
n→ n 13
an
n3
2n
2n3
2
n→ n 13
lim
Therefore, by the Ratio Test, the series diverges.
52.
1
35.
. . 2n 1
2 5 8 . . . 3n 1
n1
lim
n→ an1
1
lim
n→ 2
an
3.
5.
By the Ratio Test, the series diverges.
2n1
01 0 < 1
51.
en
n!
. . 2n 12n 1
. . 3n 13n 2
By the Ratio Test, the series converges.
2
1
5.
3.
. . 3n 1
2n 1 2
lim
. . 2n 1 n→
3
3n 2
Review Exercises for Chapter 9
53. (a) Ratio Test: lim
n→ (b)
an1
n 135n1
n1
lim
lim
n→ n→ an
n35n
n
x
5
10
15
20
25
Sn
2.8752
3.6366
3.7377
3.7488
3.7499
35 53 < 1,
(c)
Converges
4
0
(d) The sum is approximately 3.75.
12
−1
54. (a) The series converges by the Alternating Series Test.
(b)
x
5
10
15
20
25
Sn
0.0871
0.0669
0.0734
0.0702
0.0721
(c)
(d) The sum is approximately 0.0714.
0.3
0
12
0
55. (a)
1
N
x2
1x
dx N
1
N
N
5
10
20
30
40
n
1.4636
1.5498
1.5962
1.6122
1.6202
0.2000
0.1000
0.0500
0.0333
0.0250
5
10
20
30
40
n
1.0367
1.0369
1.0369
1.0369
1.0369
0.0004
0.0000
0.0000
0.0000
0.0000
N
1
2
n1
1
dx
x2
N
(b)
N
1
1
dx 4
x5
4x
N
1
4N4
N
N
1
5
n1
1
dx
x5
N
The series in part (b) converges more rapidly. The integral values represent the remainders of the partial sums.
56. No. Let an 3937.5
3937.5
, then a75 0.7. The series
is a convergent p-series.
n2
n2
n1
57. f x ex2
1
fx ex2
2
1 x2
f x e
4
1
fx ex2
8
f 0 1
f0 f 0 x2
x3
f 0
2!
3!
1
1 x2
1 x3
1 x
2
4 2! 8 3!
1
1
1
1 x x2 x3
2
8
48
P3x f 0 f0x f 0
1
2
1
4
f0 58. f x tan x
1
8
4 1
f 4 2
fx sec2 x
f f x 2 sec2 x tan x
f fx 4 sec2 x tan2 x 2 sec4 x
f 4 4
4 16
P3x 1 2 x 2 x
4
4
2
8
x
3
4
3
347
348
Chapter 9
Infinite Series
959
< 0.001, use four terms.
1809 9!
59. Since
sin 95
sin
60. cos0.75 1 95
95
95
0.99594
95180 95180 180
3! 180 5! 180 7!
3
5
3
7
5
7
0.752 0.754 0.756
0.7317
2!
4!
6!
61. ln1.75 0.75 0.752 0.753 0.754 0.755 0.756 . . . 0.7514
0.559062
2
3
4
5
6
14
0.252 0.253 0.254
0.779
2!
3!
4!
62. e0.25 1 0.25 63. f x cos x, c 0
Rnx f n1z
n 1!
f n1z
64.
x n1
≤ 1 ⇒ Rnx ≤
(a) Rnx ≤
x n1
n 1!
0.5n1
< 0.001
n 1!
f x cos x
P4x 1 x2
x4
2! 4!
P6x 1 x2
x4
x6
2! 4! 6!
P10x 1 This inequality is true for n 4.
(b) Rnx ≤
10
P4
1n1
< 0.001
n 1!
−10
10
f
This inequality is true for n 6.
(c) Rnx ≤
x4
x6
x8
x10
x2
2! 4! 6! 8! 10!
P6
P10
−10
0.5n1
< 0.0001
n 1!
This inequality is true for n 5.
(d) Rnx ≤
2n1
< 0.0001
n 1!
This inequality is true for n 10.
65.
10
x
n
66.
Geometric series which converges only if 2x < 1 or
1
1
2 < x < 2 .
Geometric series which converges only if x10 < 1 or
10 < x < 10.
1nx 2n
n 12
n0
lim
n→ n
n0
n0
67.
2x
68.
un1
1n1x 2n1
lim
n→ un
n 22
n 12
1nx 2n
x2
R1
Center: 2
Since the series converges when x 1 and when x 3,
the interval of convergence is 1 ≤ x ≤ 3.
3nx 2n
n
n1
lim
n→ un1
3n1x 2n1
lim
n→ un
n1
n
3nx 2n
3x2
1
3
Center: 2
R
5
7
Since the series converges at 3 and diverges at 3 , the
interval of convergence is 53 ≤ x < 73 .
Review Exercises for Chapter 9
69.
n!x 2
n
n0
lim
n→ 70.
un1
n 1!x 2n1
lim
n→ un
n!x 2n
x2
x 2n
n
2
2
n0
n0
y
x2
< 1 or
2
n
y 1n12n 2x2n1
1n2nx2n1
n
2
4 n!
4n1
n 1! 2
n1
n0
y 1n12n 22n 1x2n
4n1
n 1! 2
n0
1n12n 2x2n2
1n12n 22n 1x2n2
x2n2
1n n 2
n1
n 1! 2
n1
n 1! 2
4
4
4
n!
n0
n0
n0
1
n1
y
1n14n 12
1
1n n
x 2n2
4n1
n 1! 2
4 n! 2
1n11
1
1n n
x 2n2 0
4nn! 2
4 n! 2
3nx 2n
2nn!
n0
3n12n 2x2n1
3n2nx2n1
n
2 n!
2n1n 1!
n1
n0
y 3n12n 22n 1x2n
2n1n 1!
n0
n0
2 x
n
1n13n22n 2x2n2
1n3n1x2n
3n12n 22n 1x2n
n1
n1
2
n
1
!
2
n
1
!
2nn!
n0
n0
n0
1n13n2x2n2
1n3n1x2n
1n13n12n 2x 2n
nn!
nn!
2
2
2nn!
n0
n0
n0
1n13n2x 2n2
1n3n1x2n
2n 1 1 n
2 n!
2nn!
n0
n0
1n13n2x 2n2
1n3n1x2n
2n n
2 n!
2nn!
n0
n0
1n3n1x 2n 2n
1n13n1x2n
2n nn!
n1n 1! 2n
2
n1
n1 2
1n3n1x2n
2n 2n 0
2nn!
n1
n0
2
23
a
3 x 1 x3 1 r
3 3
y y 3xy 3y 73.
n0
72.
2n 22n 1 1n12n 2
1n
n1
n 2 x 2n2
4n1
n 1! 2
4 n 1! 2
4 n!
1n12n 22n 1 1
1
1n n
x 2n2
4n1
n 1! 2
4 n! 2
n0
n0
0 < x < 4.
n0
n
x2n
4nn!2
1
n0
x2y xy x2y Geometric series which converges only if
which implies that the series converges only at the center
x 2.
71.
2x n
3n1
74.
3
32
32
a
2 x 1 x2 1 x2 1 r
2 2 n0
3
x
n
1n3xn
n1
n0 2
349
350
Chapter 9
75. gx Infinite Series
2 x
2
. Power series
3x
n0 3 3
Derivative:
n
76. Integral:
x
n1
1
n1
1n3xn1
n1
n0
3 n 3 3 9 n 3 2
n 12
2
x
n1
n1
9 n 13
2
x
n
n0
2x
2
4
8
77. 1 x x2 x3 . . . 3
9
27
n0 3
n
1
3
,
1 2x3 3 2x
1
1
x 3
78. 8 2x 3 x 32 x 33 . . . 8
2
8
4
n0
79.
n
3
3
< x <
2
2
8
1 x 34
32
32
, 1 < x < 7
4 x 3 1 x
f x sin x
fx cos x
f x sin x
fx cos x, . . .
sinx 80.
f nx
x 34 n
n!
n0
2
2
3
3
2
x
x
2 4 2 2! 4
2
2
2 1nn12
x 34 n
. . .
2 n0
n!
f x cos x
fx sin x
f x cos x
fx sin x
cos x 2
f n 4
x 4 n 2 2
x
x
n!
2
2
4
2 2!
4
n0
1
1 x 2 4
2
4 n1
n 1!
n1
xn
n!, we have 3
x
f 4x 5 csc5x 15 csc3x cot2x cscx cot4x
2
x 2 4! 4
2
4
. . .
5
x
4!
2
fx 5 csc3x cotx cscx cot3x
1
f n2
x 2 n
1
x
n!
2!
2
n0
3
f x csc3x cscx cot2x
x 2 3! 4
2
x2 ln 32 x3 ln 33 x 4 ln 34 . . .
x ln 3n
1 x ln 3 .
n!
2!
3!
4!
n0
fx cscx cotx
f x cscx
cscx 2
n0
82.
nn12
x
81. 3x eln3x ex ln3 and since ex 4
. . .
Review Exercises for Chapter 9
83.
f x 1
x
fx f x 1
x2
2
x3
6
fx 4, . . .
x
f n1x 1 n
n!x 1n
1
x 1n, 2 < x < 0
x n0
n!
n!
n0
n0
84.
f x x12
1
fx x12
2
1 1 32
f x x
2 2
f x 121232x
52
f 4x x 12123252x
72,
. . .
f n4x 4n
x 4 x 42 1
2
n!
22
252!
n0
2
3x 43 1 3 5x 44 .
283!
2114!
. .
1n11 3 5 . . . 2n 3x 4n
x 4
2
2
23n1n!
n2
85. 1 xk 1 kx 1 x15 1 kk 1x2 kk 1k 2x3 . . .
2!
3!
x
1545x2 154595x3 . . .
5
2!
3!
1
1 4x2 1 4 9x3 . . .
1 x 2
5
5 2!
533!
86.
9 14 .
1
1n14
x
5 n2
1
x
2
6 3 . . .
x2 x 5 25
125
. . 5n 6xn
5nn!
hx 1 x3
hx 31 x4
hx 121 x5
hx 601 x6
h4x 3601 x7
h5x 25201 x8
1nn 2!x n
1nn 2n 1xn
1
12x2 60x3 360x4 2520x5 . . .
1 3x 1 x3
2!
3!
4!
5!
2n!
2
n0
n0
351
352
Chapter 9
1
ln x 87.
Infinite Series
n1
n1
ln
x 1n
,
n
0 < x ≤ 2
n
n1
ln
1
n1
n1
89.
ex 1
0.2231
4nn
91.
90.
1n
n0
cos
23 1
n
n0
x2n
,
2n!
n1
xn
n!,
ex n0
12n
1
1.6487
n
n!
n0
n0 2 n!
cos x n1
xn
n
n1
1
n!, < x < 0 < x ≤ 2
65 1 65n 1
n0
e12 x 1n
,
n
n1
n1
n1
n1
54 1 54n 1
1
ln x 88.
e23 < x <
92.
< x <
22n
0.7859
3 2n!
sin x sin
2n
23n
1.9477
n
n!
n0
n0 3 n!
1n
n0
2n
1
0.1823
5nn
13 1
n
n0
x2n1
,
2n 1!
< x <
1
0.3272
32n12n 1!
93. The series for Exercise 41 converges very slowly because the terms approach 0 at a slow rate.
94.
1
1 x 3
1 x 312, k 1
2
1
1232 6 123252 9 . . .
1 x 3 x x 2
2!
3!
1
1n1
x3
2
n2
95. (a) f x e 2x
96. (a)
3.
. . 2n 1
2nn!
x 3n
f 0 1
(b)
f0 2
f x 4e 2x
f 0 4
f x 8e 2x
f 0 8
Px 1 2x 4x 2 8x 3
4
1 2x 2x 2 x 3
2!
3!
3
f 0 0
fx 2 cos 2x
f0 2
f x 4 sin 2x
f 0 0
fx 8 cos 2x
f0 8
2x
2xn
n0 n!
(c) e x e x 1 x x2 . . .
2!
1 x 2!x . . .
4
P 1 2x 2x 2 x 3
3
f 40 0
16 sin 2x
f 5x 32 cos 2x
f 50 32
f 6x 64 sin 2x
f 60 0
f 7x 128 cos 2x
f 70 128
—CONTINUED—
xn
4
Px 1 2x 2x 2 x 3
3
f x sin 2x
sin 2x 0 2x n!, e
n0
fx 2e2x
f 4x
ex 4
0x2 8x3 0x4 32x5 0x6 128x7 . . .
4
8 7 . . .
2x x3 x5 x 2!
3!
4!
5!
6!
7!
3
15
315
2
Review Exercises for Chapter 9
96. —CONTINUED—
(b) sin x sin 2x 1nx2n1
n0 2n 1!
1n2x2n1
2x3 2x5 2x7 . . .
2x 2n 1!
3!
5!
7!
n0
2x 4
4
8 7 . . .
8x3 32x5 128x7 . . .
2x x3 x5 x 6
120
5040
3
15
315
(c) sin 2x 2 sin x cos x
x5
x7
x3
. . .
6
120 5040
4 x7
2x3 2x5
4
4
8 7 . . .
. . . 2x x3 x5 x 3
15
315
3
15
315
2 x
2 x 2 x
sin t 97.
x
. . .
1 x2 24x 720
2
4
6
1nt 2n1
n0 2n 1!
0
sin t
dt t
cos t 98.
1n t 2n
sin t
t
n0 2n 1!
x
cos
1n t 2n1
n0 2n 12n 1!
x
x
cos
0
0
1
nx 2n1
2n 12n 1!
t
t
2
2
dt 1
1n t n
1 t n0
ln1 t x
0
1
1
dt 1t
n1
n0
1n t n1
2
n0 n 1
arctan x x et 1 lim
1n t n1
2
2n!n 1
n0
2n
0
1 x
2n!n 1
n n1
2
x
2n
tn
n!
x
0
tn
n1
1n x n1
2
n0 n 1
x
0
et 1
dt t
x3 x5 x7 x9 . . .
3
5
7
9
arctan x
0
x
n!
tn1
et 1
t
n1 n!
x 52 x 92 x132 x172 . . .
arctan x
x 3
5
7
9
x
x→0 1n t n
2n2n!
n0
2
n0
n t n1
101.
et 100.
1n t n
lnt 1
t
n0 n 1
lnt 1
dt t
1n t 2n
2n!
n0
n0
n0
99.
x5
x5
x5
x7
x7
x7
x3 x3
x7
. . .
2
6
24 12 120
720 144 240 5040
1
arctan x
1
x2
2x
By L’Hôpital’s Rule, lim
lim
lim
0.
x→0
x→0
1
x
x→0 1 x2
2x
tn
x
n n!
n1
0
xn
n n!
n1
353
354
Chapter 9
Infinite Series
arcsin x x 102.
x3
2
3
1 3x5
1 3 5x7
. . .
245 2467
x2
1 3x4
1 3 5x6
arcsin x
1
. . .
23 245 2467
x
lim
x→0
arcsin x
1
x
1
arcsin x
1 x2
By L’Hôpital’s Rule, lim
lim
1.
x→0
x→0
x
1
Problem Solving for Chapter 9
1. (a) 1
3 29 427 . . . 3 3
1
1
1
1 2
n
n0
2. Let S n→ n0
1 2
3 3
n
13
1
1 23
Then
110
1
1
. . .
22 42
S
1 2
1
1
1
1 2 2. . . S 2
.
22
2
3
2 6
a1 1 and r1 1 3
3 2
3
6
1
.
23
r1
1
2
a2 3 and 1 23r2 2r2
1
.
2 23
2r2
1
r2
3 r2
For six circles,
a3 6 and 1 23r3 4r3
r3 1
.
23 4
2r3
1
Continuing this pattern, rn Total Area rn2an An lim An n→ 2
1
4
8
1
23 2n 1
.
23 12n 1
2
nn 1
2 23 2n 1
2
nn 1
2
For three circles,
r2 2 1 2 2 3
2
.
6
4 6
6 4
8
3
2
1
1
1
1
. . ..
12 32 52
S
nn 1
.
2
For one circle,
1
1
1
2
1
2 2 2 2. . .
6
1
2
3
4
Thus, S 3. If there are n rows, then an 2
n1
1 2
(b) 0, , , 1, etc.
3 3
(c) lim Cn 1 1
2n 1
3 r3
r3
Problem Solving for Chapter 9
4. (a) Position the three blocks as indicated in the figure. The bottom block extends 16 over
the edge of the table, the middle block extends 14 over the edge of the bottom block,
and the top block extends 12 over the edge of the middle block.
1
6
The centers of gravity are located at
bottom block:
1 1
1
6 2
3
middle block:
1 1 1
1
6 4 2
12
top block:
0 1 5
6 12
1 1 1 1
5
.
6 4 2 2 12
The center of gravity of the top 2 blocks is
121 125 2 61,
which lies over the bottom block. The center of gravity of the 3 blocks is
31 121 125 3 0
which lies over the table. Hence, the far edge of the top block lies
1 1 1 11
6 4 2 12
beyond the edge of the table.
n
(b) Yes. If there are n blocks, then the edge of the top block lies
1
2i from the
i1
edge of the table. Using 4 blocks,
4
1
1
1
1
1
25
2i 2 4 6 8 24
i1
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 .
i1
5. (a)
a x
n
n
i1
1 2x 3x2 x3 2x4 3x5 . . .
1 x3 x6 . . . 2x x4 x7 . . . 3x2 x5 x8 . . .
1 x3 x6 . . .
1 2x 3x2
1 2x 3x2
1
1 x3
R 1 because each series in the second line has R 1.
(b)
a x
n
n
a0 a1x . . . ap1x p1 a0 x p a1x p1 . . . . . .
a01 x p . . . a1 x1 x p . . . . . . a p1 x p11 x p . . .
a0 a1 x . . . a p1x p11 x p . . .
a0 a1 x . . . ap1 x p1
R1
Assume all an > 0.
1
2
1
4
1
1 xp
11
12
355
356
Chapter 9
6. a Infinite Series
1n1a b a b
b a b . . .
2 3 4
2n
n1
If a b,
1n1
1n12a
a
converges conditionally.
2n
n
n1
n1
If a b,
ab
1n1a b
diverges.
2n
n1
n1 2n
No values of a and b give absolute convergence. a b implies conditional convergence.
ex 1 x 7. (a)
xe x xn
x2 . . .
2!
n0 n!
ex 1 x 8.
x n1
n0 n!
ex 1 x2 2
xe x dx xe x e x C x n2
n0 n 2n!
1
1
1
n 2n! 2 n 2n!.
n0
n1
1
1
Thus,
.
n
2
n!
2
n1
(b) Differentiating,
xe x e x n 1xn
.
n!
n0
Letting x 1,
2e 9. Let a1 0
n1
5.4366.
n!
n0
sin x
dx, a2 x
2
sin x
dx, a3 x
3
2
sin x
dx, etc.
x
Then,
0
sin x
dx a1 a2 a3 a4 . . . .
x
Since lim an 0 and an1 < an, this series converges.
n→ 10. (a) If p 1,
2
If p > 1,
2
1
dx ln ln x
x ln x
diverges.
2
1
ln b1p ln 21p
converges.
dx lim
b→ xln xp
1p
1p
If p < 1, diverges.
(b)
n4
1
1
1
diverges by part (a).
2
n lnn 2n4 n ln n
x4 . . . x12 . . .
2!
6!
12!
f 120
1
⇒ f 120 665,280
12!
6!
6!
Letting x 0, you have C 1. Letting x 1,
ee1
x2 . . .
2!
Problem Solving for Chapter 9
11. (a) a1 3.0
a2 1.73205
a3 2.17533
a4 2.27493
a5 2.29672
a6 2.30146
lim an n→ 1 13
[See part (b) for proof.]
2
(b) Use mathematical induction to show the sequence is increasing. Clearly, a2 a a1 aa > a a1.
Now assume an > an1. Then
an a > an1 a
an a > an1 a
an1 > 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
aa 1 > an1
a2 a > an
a2 > an a
a > an a an1.
Hence, the sequence converges to some number L. To find L, assume an1 an L:
L a L ⇒ L2 a L ⇒ L2 L a 0
L
1 ± 1 4a
.
2
Hence, L 1 1 4a
.
2
12. Let bn an r n.
bn 1n an r n1n an1n r → Lr as n → .
1
Lr < r 1.
r
By the Root Test,
13. (a)
1
2
n1
n 1n
S1 b
n
converges ⇒
a r
n
n
converges.
1
1
1
1
1
. . .
211 221 231 241 251
1
1
20
S1 1 1 9
8 8
S3 9 1 11
8 4
8
S4 1
45
11
8
32 32
S5 45
1
47
32 16 32
—CONTINUED—
357
358
Chapter 9
Infinite Series
13. —CONTINUED—
2n 1
21
an1
n1 1n 1 1 1 n 1
an
2
2
n
(b)
n
1
1
This sequence is 8, 2, 8, 2, . . . which diverges.
(c)
2
n
1
n 1n
14. (a)
2
1n
1
1n
2
1
1
n
1
1
n 12 → 1 and n 2 → 1.
→
< 1 converges because 21 2, 2, 2, 2, . . . and n 1 n
2
2
n
2
1
1
0.01n
0.99 1 0.01 n0
15. S6 130 70 40 240
1 0.01 0.012 . . .
1.010101 . . .
(b)
S10 1490 810 440 2740
1 0.02 0.022 . . .
1 0.02 0.0004 . . .
1.0204081632 . . .
1
2
2
1
n
12
n1
(b) S 4 1 1
3
. . .
p-series, p 21 < 1
1
1 1 . . .
4
2 3
n1 n
4
1
1
(c) W 1 32 32 . . .
3
2
3
4 1
converges.
3 n1 n32
17. (a)
1
1
1
2n 2 n diverges (harmonic)
n1
n1
(b) Let f x sin x. By the Mean Value Theorem,
f x f y fcx y coscx y ≤ x y,
where c is between x and y. Thus,
0 ≤ sin
2n1 sin2n 1 1
1
1
≤
2n 2n 1
1
2n2n 1
Because
1
2n2n 1 converges, the Comparison Theorem
n1
tells us that
sin2n sin2n 1 converges.
n1
1
S8 440 240 130 810
S9 810 440 240 1490
1
1
0.02n
0.98 1 0.02 n0
16. (a) Height 2 1 S7 240 130 70 440
1