Solução_Calculo_Stewart_6e

F.
476 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
7. lim
k→∞
a
k+1 (k +1) 2
k+1
3
= lim
k→∞
∞
n=1
a k
k
2 k
= lim
k→∞
3
k +1
k
2
3
1
= 2
3 lim 1+ 1
= 2
k→∞
3
k
(1) = 2 < 1,sotheseries
3
k 2
3
k
is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the
same as convergence.
9. lim
n→∞
a n+1
a n
(1.1)
n+1
= lim
n→∞ (n +1) 4 ·
=(1.1)(1) = 1.1 > 1,
n 4
= lim
(1.1) n n→∞
TX.10
(1.1)n 4
4
=(1.1) lim
(n +1) n→∞
1
(n +1) 4
n 4
1
=(1.1) lim
n→∞ (1 + 1/n) 4
so the series ∞ (−1) n (1.1) n
diverges by the Ratio Test.
n=1
11. Since 0 ≤ e1/n
n 3
n 4
≤ e n 3 = e 1
n 3
and ∞
n=1
1
n is a convergent p-series [p =3> 1], ∞
3
n=1
e 1/n
n 3
converges, and so
∞
n=1
13. lim
n→∞
(−1) n e 1/n
is absolutely convergent.
n 3
a n+1
a n
= lim
n→∞
10 n+1 (n +1)42n+1
·
(n +2)42n+3 10 n
10
= lim
n→∞ 4 · n +1
= 5 2 n +2 8 < 1,sotheseries ∞
n=1
10 n
(n +1)4 2n+1
is absolutely convergent by the Ratio Test. Since the terms of this series are positive, absolute convergence is the same as
convergence.
15.
(−1)n arctan n
< π/2
n 2 n ,sosince ∞
2
17.
19.
n=1
converges absolutely by the Comparison Test.
∞
n=2
(−1) n
ln n
π/2
n 2 = π 2
∞
n=1
converges by the Alternating Series Test since lim
n→∞
1
ln n > 1 n , and since ∞
∞
n=2
(−1) n
ln n
|cos (nπ/3)|
n!
Comparison Test.
n=2
1
n is the divergent (partial) harmonic series, ∞
is conditionally convergent.
≤ 1 n! and ∞
n=1
1
n converges (p =2> 1), the given series ∞
2
n=1
(−1) n arctan n
n 2
1
1
ln n =0and is decreasing. Now ln n1 (byEquation3.6.6),sotheseries ∞ 1+ 1 n
2
n→∞
n→∞ n n→∞ n
n=1 n
diverges by the Root Test.