Solução_Calculo_Stewart_6e

F.
480 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
29.
∞
a n = ∞ (−1) n 1
cosh n = ∞ (−1) n b n .Nowb n = 1
cosh n > 0, {b n} is decreasing, and lim b n =0,sotheseries
n→∞
n=1
n=1
n=1
converges by the Alternating Series Test.
Or: Write
1
cosh n = 2
e n + e −n < 2
e n and ∞
n=1
1
e is a convergent geometric series, so ∞
n
Comparison Test. So ∞ (−1) n 1
is absolutely convergent and therefore convergent.
cosh n
31. lim
k→∞ a k = lim
k→∞
Thus,
∞
k=1
n=1
5 k
3 k +4 k =[divide by 4k ] lim
k→∞
5 k
diverges by the Test for Divergence.
3 k +4k n=1
1
is convergent by the
cosh n
(5/4) k
k k 3 5
= ∞ since lim =0and lim = ∞.
(3/4) k +1 k→∞
4
k→∞
4
33. Let a n = sin(1/n) √ and b n = 1
n n √ a n sin(1/n)
.Then lim = lim =1> 0, so ∞ sin(1/n)
√ converges by limit
n n→∞ b n n→∞ 1/n
n=1 n
comparison with the convergent p-series ∞ 1
[p =3/2 > 1].
n 3/2
n
35. lim |an| = lim
n→∞
n→∞
n
n +1
converges by the Root Test.
n=1
n
2 /n
= lim
n→∞
1
[(n +1)/n] n =
1
lim (1 + = 1
n→∞ 1/n)n e < 1,sotheseries ∞
n=1
n
37. lim |an| = lim
n→∞
n→∞ (21/n − 1) = 1 − 1=0< 1,sotheseries ∞ √ n
n
2 − 1 converges by the Root Test.
n=1
n
n +1
n
2
11.8 Power Series
1. A power series is a series of the form ∞
n=0 c nx n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + ···,wherex isavariableandthec n ’s are
constants called the coefficients of the series.
More generally, a series of the form ∞
n=0 c n(x − a) n = c 0 + c 1 (x − a)+c 2 (x − a) 2 + ··· is called a power series in
(x − a) or a power series centered at a or a power series about a,wherea is a constant.
3. If a n = xn
√ n
,then lim
n→∞
a n+1 = lim √ · n +1
a n
By the Ratio Test, the series ∞
n=1
n→∞ xn+1
endpoints, that is, x = ±1. Whenx =1,theseries ∞
the series ∞
n=1
5. If a n = (−1)n−1 x n
,then
a n
√ n
= lim
x n n→∞
x
√ √ = lim
n +1/ n n→∞
|x|
1+1/n
= |x|.
x n
√ n
converges when |x| < 1, so the radius of convergence R =1. Now we’ll check the
n=1
1
√ diverges because it is a p-series with p = 1 ≤ 1. Whenx = −1,
n
2
(−1) n
√ n
converges by the Alternating Series Test. Thus, the interval of convergence is I =[−1, 1).
n 3
lim
n→∞ a
n+1 = lim
n→∞ (−1)n x n+1 n 3 3
·
= lim
(n +1) 3 (−1) n−1 x n n→∞ (−1)xn3
n
= lim
|x|
=1 3 ·|x| = |x|. Bythe
(n +1) 3 n→∞ n +1