SECTION 8.1
8.1
■
List the first five terms of the sequence.
1 n
3. a n 2n
2
4. a n 3
25.
n
27.
n
sin
2
1
1 an
1 3 5 2n 1
7. a n n!
8.
7 n1
n!
■
■
■
■
■
■
■
■
■
9. 1, 4, 7, 10, . . .
10.
11.
{32 , 94 , 278 , 8116 , . . .}
13.
{ ,
■
, , , . . .}
■
■
■
■
1
15. a n 4n 2
16. a n 4 sn
n2 1
17. a n 2
n 1
4n 3
18. a n 3n 4
20. a n n2
1 n3
22.
1 n sin
1
n
ln2 e n
3n
32. a n 1 3n1
n
37.
■
■
3
4
ns
n
s
5
n
sn s
n
3n
arctan
1
2
n4
1 n2 n3
2n
2n 1
sin n
sn
■
38. a n 1
2
n
2 2
n2
n
n
39. a n n cos n
n2 1
■
n2
n1
31. a n n2 n
■
■
■
■
■
■
■
■
■
■
■
■
40 – 43 ■ Determine whether the sequence is increasing,
decreasing, or not monotonic. Is the sequence bounded?
40. a n 1
3n 5
41. a n n2
n2
42. a n 3n 4
2n 5
43. a n sn
n2
■
Copyright © 2013, Cengage Learning. All rights reserved.
28.
n!
n 2!
30.
36.
14. 0, 2, 0, 2, 0, 2, . . .
■
■
21. a n 1 n
lnn 2 n
29. sn 2 sn 12. 1, 2, 6, 24, . . .
5
9
■
19. a n 26.
35. a n 1 n1
{163 , 254 , 365 , 496 , . . .}
Determine whether the sequence converges or diverges.
If it converges, find the limit.
15–39
3 1 n
n2
■
■ Find a formula for the general term a of the sequence,
n
assuming that the pattern of the first few terms continues.
■
34. a n (sn 1 sn ) sn 9–14
3 4
5 7
24. a n 2 cos n
33. a n n 1
n
■
2
3
n
2
23. a n sin
4n 3
2. a n 3n 4
n1
6. a 1 1, a n1 1
S Click here for solutions.
n
1. a n 2n 1
5.
■
SEQUENCES
A Click here for answers.
1– 8
SEQUENCES
■
■
■
■
■
■
■
■
2
■
SECTION 8.1
SEQUENCES
8.1
ANSWERS
E Click here for exercises.
1.
2.
3.
4.
5.
1 2 3 4 5
, , , ,
3 5 7 9 11
1 1 9 13 17
, , , ,
7 2 13 16 19
1
1 3
1 5
,− , ,− ,
2
2 8
4 32
2 4 −8 16
32
− , ,
, ,−
3 9 27 81
243
1, 0, −1, 0, 1
1
6. 1, ,
2
3
7. 1, ,
2
2 3 5
, ,
3 5 8
5 35 63
, ,
2 8 8
343 2401
16,807 117,649
8. 49, −
,
,−
,
2
6
24
120
9. an = 3n − 2
10. an =
n
3
n+1
11. an = (−1)
2
n
n+1
2n + 1
n+1
n−1
14. an = 1 − (−1)
22. Diverges
23. Diverges
24. Diverges
25. 0
26. 0
27. 0
28. 0
29. 0
1
3
31. 0
30.
33. 1
34.
or an = 1 + (−1)n
36.
37.
38.
1
2
Diverges
π
4
0
1
2
0
16. Diverges
39.
17. 1
40. Decreasing; yes
4
3
19. Diverges
41. Increasing; yes
18.
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21. 0
35.
n!
13. an = (−1)
15. 0
20. 0
32. 1
n+2
(n + 3)2
12. an = (−1)
S Click here for solutions.
42. Increasing; yes
43. Not monotonic; yes
SECTION 8.1
8.1
SEQUENCES
■
SOLUTIONS
E Click here for exercises.
1. an =
n
, so the sequence is
2n + 1
2. an =
4n − 3
, so the sequence is
3n + 4
1 2 3 4 5
, , , , ,... .
3 5 7 9 11
1 1 9 13 17
, , , , ,... .
7 2 13 16 19
(−1)n−1 n
, so the sequence is
2n
1
1 3
1 5
,− , ,− , ,... .
2
2 8
4 32
n
2
4. an = −
, so the sequence is
3
2 4 −8 16
32
− , ,
, ,−
,... .
3 9 27 81
243
nπ
5. an = sin
, so the sequence is {1, 0, −1, 0, 1, . . .}.
2
1
6. a1 = 1, an+1 =
, so the sequence is
1 + an
1
1
1
1
,
,
,
.
.
.
,
1,
1 + 1 1 + 12 1 + 23 1 + 35
1 1 1 1
1 2 3 5
= 1, , 3 , 5 , 8 = 1, , , , , . . . .
2 2 3 5
2 3 5 8
3. an =
1 · 3 · 5 · · · · · (2n − 1)
7. an =
, so the sequence is
n! 3 5 35 63
1, , , , , . . . .
2 2 8 8
(−7)n+1
, so the sequence is
n!
16,807 117,649
343 2401
,
,−
,
,... .
49, −
2
6
24
120
8. an =
9. an = 3n − 2
10. an =
n+2
(n + 3)2
n+1
11. an = (−1)
n
12. an = (−1)
n!
n+1
13. an = (−1)
n
3
2
n+1
2n + 1
14. {0, 2, 0, 2, 0, 2, . . .}. 1 is halfway between 0 and 2, so we can
Copyright © 2013, Cengage Learning. All rights reserved.
think of alternately subtracting and adding 1 (from 1 and to
1) to obtain the given sequence: an = 1 − (−1)n−1 .
15. lim
n→∞
1
1
1
=
=
lim
4n2
4 n→∞ n2
1
4
· 0 = 0. Convergent
√
√
n} clearly diverges since n → ∞ as n → ∞.
16. {4
n2 − 1
1 − 1/n2
= 1. Convergent
= lim
2
n→∞ n + 1
n→∞ 1 + 1/n2
17. lim
18. lim
n→∞
4n − 3
4 − 3/n
4
= lim
= . Convergent
3n + 4 n→∞ 3 + 4/n
3
19. {an } diverges since
n2
n
=
→ ∞ as n → ∞.
n+1
1 + 1/n
1/n1/6 + 1/n1/4
n1/3 + n1/4
0
= lim
= = 0 so
1/2
1/5
n→∞ n
n→∞
1
+n
1 + 1/n3/10
the sequence converges.
20. lim
1/n
n2
= lim
= 0, so by
n→∞
n→∞ 1 + n3
n→∞ (1/n3 ) + 1
n2
= 0.
Theorem 6, lim (−1)n
n→∞
1 + n3
n
22. an = π
, so {an } diverges by Equation 8 with
3
π
r = 3 > 1.
21. lim |an | = lim
23. {an } = {1, 0, −1, 0, 1, 0, −1, . . .}. This sequence oscillates
among 1, 0, and −1, so the sequence diverges.
24. an = 2 + cos nπ, so
{an } = {2 + cos π, 2 + cos 2π, 2 + cos 3π, 2 + cos 4π, . . .}
= {2 − 1, 2 + 1, 2 − 1, 2 + 1, . . .}
= {1, 3, 1, 3, . . .}
This sequence oscillates between 1 and 3, so it diverges.
3 + (−1)n
3 + (−1)n
4
4
25. 0 <
≤
and
lim
=
0,
so
n→∞ n2
n2
n2
n2
converges to 0 by the Squeeze Theorem.
n!
1 · 2 · 3 · ··· · n
= lim
n→∞ 1 · 2 · 3 · · · · · n (n + 1) (n + 2)
(n + 2)!
1
= lim
=0
n→∞ (n + 2) (n + 1)
Convergent
ln x2
2 ln x H
2/x
27. lim
= lim
= lim
= 0, so by
x→∞
x→∞
x→∞ 1
x
x
2
ln n
converges to 0.
Theorem 3,
n
26. lim
n→∞
1
1
28. lim sin
= sin 0 = 0 since → 0 as n → ∞, so by
n→∞
n
n
1
converges to 0.
Theorem 6, (−1)n sin
n
√
√
√
√
√
√ n+2+ n
29. bn = n + 2 − n =
n+2− n √
√
n+2+ n
1
2
2
= √
√ < √ = √ → 0 as n → ∞.
2 n
n
n+2+ n
√
So by the Squeeze Theorem with an = 0 and cn = 1/ n,
√
√ n + 2 − n converges to 0.
3
4
■
SECTION 8.1
SEQUENCES
ln (2 + ex ) H
ex / (2 + ex )
= lim
x→∞
x→∞
3x
3
1
1
= lim
=
x→∞ 6e−x + 3
3
ln (2 + en )
1
so by Theorem 3, lim
= . Convergent
n→∞
3n
3
x H
1
31. lim x = lim
= 0, so by Theorem 3, n2−n
x→∞ 2
x→∞ (ln 2) 2x
converges to 0.
30. lim
1
ln (1 + 3x) ⇒
x
ln (1 + 3x) H
3/ (1 + 3x)
= lim
=0
lim ln y = lim
x→∞
x→∞
x→∞
x
1
⇒ lim y = e0 = 1, so by Theorem 3, (1 + 3n)1/n
1/x
32. y = (1 + 3x)
⇒ ln (y) =
x→∞
converges to 1.
ln x
. Then ln y = −
and
x
1/x
H
= 0, so lim y = e0 = 1,
lim (ln y) = lim −
x→∞
x→∞
x→∞
1
and so {an } converges to 1.
√
√ 34. an =
n+1− n
n + 12
√
√ √
√ n+1+ n
√
=
n+1− n
n + 12
√
n+1+ n
n + 1/2
1 + 1/ (2n)
1
→ as n → ∞.
= √
√ = 2
n+1+ n
1 + 1/n + 1
Convergent
n
35. |an | =
→ ∞ as n → ∞, so {an }
1/n3 + 1/n + 1
diverges.
−1/x
33. Let y = x
2n
2
= lim
= 1, so
n→∞ 2 + 1/n
2n + 1
2n
= arctan 1 = π4 . Convergent.
lim arctan
n→∞
2n + 1
36. lim
n→∞
1
|sin n|
√
≤ √ → 0 as n → ∞, so by the Squeeze
n
n
sin n
Theorem and Theorem 6, √
converges to 0.
n
37. 0 <
38. The series converges, since
1 + 2 + 3 + ··· + n
n2
n (n + 1) /2
=
[sum of the first n positive integers]
n2
1 + 1/n
1
n+1
=
→ as n → ∞.
=
2n
2
2
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an =
n |cos n|
n
1
≤ 2
=
→ 0 as
n2 + 1
n +1
n + 1/n
n → ∞, so by the Squeeze Theorem and Theorem 6, {an }
converges to 0.
39. 0 ≤ |an | =
1
1
<
⇔
3 (n + 1) + 5
3n + 5
an+1 < an so {an } is decreasing. The sequence is bounded
1
because an =
> 0 for n ≥ 1.
3n + 5
n−2
41.
is increasing since
n+2
n−2
(n + 1) − 2
<
an < an+1 ⇔
n+2
(n + 1) + 2
⇔ (n − 2) (n + 3) < (n + 2) (n − 1) ⇔
40. 3 (n + 1) + 5 > 3n + 5 so
n2 + n − 6 < n2 + n − 2 ⇔ −6 < −2, which is of
course true. The sequence is bounded because
n−2
n+2
<
= 1 for n ≥ 1.
n+2
n+2
3n + 4
42.
is increasing since an+1 ≥ an
2n + 5
3 (n + 1) + 4
3n + 4
⇔
≥
⇔
2 (n + 1) + 5
2n + 5
(3n + 7) (2n + 5) ≥ (3n + 4) (2n + 7) ⇔
6n2 + 29n + 35 ≥ 6n2 + 29n + 28 ⇔ 35 ≥ 28. The
3n + 4
4n + 10
sequence is bounded because an =
<
=2
2n + 5
2n + 5
for n ≥ 1.
√
n
43. an =
defines a sequence that is neither increasing nor
n+2
decreasing since a1 < a2 and a2 > a3 . (a1 = 13 = 0.3,
a2 =
√
2
4
≈ 0.354, and a3 =
√
3
5
≈ 0.346.) But the
sequence {an | n ≥ 2} obtained by omitting the first term a1
√
x
is decreasing. To see this, note that if f (x) =
for
x+2
x ≥ 0, then
x+2 √
√ − x
(x + 2) − 2x
2 x
= √
f (x) =
(x + 2)2
2 x (x + 2)2
2−x
= √
≤ 0 for x ≥ 2.
2 x (x + 2)2
The sequence is bounded since an > 0 for all n ≥ 1 and
an ≤ a2 =
√
2
4
for all n ≥ 1.