Solução_Calculo_Stewart_6e

F.
470 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
39. (a) From the figure, a 2 + a 3 + ···+ a n ≤ n
f(x) dx,sowith
1
f(x) = 1 x , 1 2 + 1 3 + 1 4 + ···+ 1 n ≤ n
1
1
dx =lnn.
x
Thus, s n =1+ 1 2 + 1 3 + 1 4 + ···+ 1 n ≤ 1+lnn.
(b) By part (a), s 10 6 ≤ 1+ln10 6 ≈ 14.82 < 15 and
s 10 9 ≤ 1+ln10 9 ≈ 21.72 < 22.
41. b ln n = e ln b ln n
= e
ln n ln b
= n
ln b = 1 .Thisisap-series, which converges for all b such that − ln b>1
n ⇔
− ln b
ln b
n
n √ n = √ 1
for all n ≥ 1,so ∞
n
p-series with p = 1 2 ≤ 1.
n=1
9 n
n
3+10 < 9n 9
n 10 = for all n ≥ 1.
n 10
converges by the Comparison Test.
cos 2 n
n 2 +1 ≤ 1
n 2 +1 < 1 n 2 ,sotheseries ∞
n=1
n=1
n
2n 3 +1 converges by comparison with ∞
n +1
n √ n diverges by comparison with ∞
∞
n=1
n=1
n=1
1
, which converges
n2 1
√
n
, which diverges because it is a
9
n
is a convergent geometric series |r| = 9 < 1
,so ∞ 9 n
10
10
n=1 3+10 n
cos 2 n
n 2 +1 convergesbycomparisonwiththep-series ∞
n − 1
n − 1
11. is positive for n>1 and
n 4n n 4 < n
n n 4 = 1 n 1 n 4 =
,so ∞ n 4
geometric series ∞ n 1
.
n=1 4
13.
arctan n
n 1.2 < π/2
n 1.2 for all n ≥ 1,so ∞
n=1
constant times a p-series with p =1.2 > 1.
n=1
n − 1
n 4 n
arctan n
n 1.2 converges by comparison with π 2
n=1
1
n 2 [p =2> 1].
converges by comparison with the convergent
∞
n=1
1
, which converges because it is a
n1.2 15.
2+(−1) n
n √ n
≤ 3
n √ n ,and ∞
n=1
3
n √ n converges because it is a constant multiple of the convergent p-series ∞
n=1
1
n √ n
p =
3
2 > 1 , so the given series converges by the Comparison Test.