Solução_Calculo_Stewart_6e

F.
TX.10
CHAPTER 11 REVIEW ¤ 505
11.
n
n 3 +1 < n n = 1 3 n ,so ∞
2
13. lim
n→∞
n=1
n
n 3 +1 converges by the Comparison Test with the convergent p-series ∞
n=1
1
[ p =2> 1].
n2 a
n+1 (n +1)
3
= lim
· 5n
= lim 1+ 1 3
· 1
a n n→∞ 5 n+1 n 3 n→∞ n 5 = 1 5 < 1,so ∞ n 3
n=1 5 convergesbytheRatioTest.
n
1
15. Let f(x) =
x √ .Thenf is continuous, positive, and decreasing on [2, ∞), so the Integral Test applies.
ln x
∞
2
so the series ∞
n=2
t
f(x) dx = lim
t→∞
2
1
x √ ln x dx u =lnx, du = 1
x dx
= lim 2 √ ln t − 2 √ ln 2
t→∞
1
n √ ln n diverges.
= ∞,
= lim
t→∞
ln t
ln 2
u −1/2 du = lim 2 √ ln t
u
t→∞ ln 2
17. |a n | =
cos 3n 1
≤
1+(1.2) n 1+(1.2) < 1 n 5 n (1.2) =
,so ∞ |a n n | converges by comparison with the convergent geometric
6 n=1
series ∞ 5
n
6 r =
5
< 1 ∞
6
. It follows that a n converges (by Theorem 3 in Section 11.6).
19. lim
n→∞
n=1
a n+1
a n
n=1
1 · 3 · 5 ·····(2n − 1)(2n +1)
= lim
·
n→∞ 5 n+1 (n +1)!
convergesbytheRatioTest.
5 n n!
1 · 3 · 5 ·····(2n − 1) = lim
n→∞
2n +1
5(n +1) = 2 5 < 1,sotheseries
√ √
n
21. b n = > 0, {bn} is decreasing, and lim
n +1 n→∞ bn =0,sotheseries ∞ n
(−1) n−1 converges by the Alternating
n=1 n +1
Series Test.
23. Consider the series of absolute values:
∞
n −1/3 is a p-series with p = 1 ≤ 1 and is therefore divergent. But if we apply the
3
n=1
Alternating Series Test, we see that b n = 3√ 1 > 0, {b n } is decreasing, and lim b
n =0,sotheseries ∞ (−1) n−1 n −1/3
n n→∞
converges. Thus,
∞
(−1) n−1 n −1/3 is conditionally convergent.
n=1
25.
a
n+1 =
a n
(−1)n+1 (n +2)3 n+1 2 2n+1
·
= n +2
2 2n+3 (−1) n (n +1)3 n n +1 · 3
4 = 1+(2/n)
1+(1/n) · 3
4 → 3 < 1 as n →∞,sobytheRatio
4
Test,
∞
n=1
(−1) n (n +1)3 n
2 2n+1 is absolutely convergent.
n=1
27.
∞
n=1
(−3) n−1
2 3n = ∞
n=1
= 1 8 ·
(−3) n−1
(2 3 ) n = ∞
8
11 = 1
11
n=1
(−3) n−1
8 n = 1 8
∞
n=1
(−3) n−1
8 n−1 = 1 8
∞
n=1
− 3 n−1
= 1
1
8 8 1 − (−3/8)