Solução_Calculo_Stewart_6e

F.
462 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
15.
17.
19.
21.
23.
∞
6(0.9) n−1 is a geometric series with first term a =6and ratio r =0.9. Since|r| =0.9 < 1, the series converges to
n=1
a
1 − r = 6
1 − 0.9 = 6
0.1 =60.
∞
n=1
(−3) n−1
converges to
∞
n=0
∞
n=1
∞
n=1
4 n = 1 4
π n
3 n+1 = 1 3
1
2n = 1 2
∞
− 3 n−1
. The latter series is geometric with a =1and ratio r = − 3
4 .Since|r| = 3 < 1, it
4 4
n=1
1
1 − (−3/4) = 4 . Thus, the given series converges to
1 4
7 4 7 =
1
. 7
∞
n=1
∞
n=0
π
3
1
n ,whichdiverges. If
n π
is a geometric series with ratio r = .Since|r| > 1, the series diverges.
3
1
n diverges since each of its partial sums is 1 times the corresponding partial sum of the harmonic series
2
∞
n=1
1
were to converge, then
∞
2n
In general, constant multiples of divergent series are divergent.
∞
k=2
n=1
1
n wouldalsohavetoconvergebyTheorem8(i).
k 2
diverges by the Test for Divergence since lim
k 2 − 1 a k 2
k = lim =16= 0.
k→∞ k→∞ k 2 − 1
25. Converges.
∞
n=1
1+2 n
3 n = ∞
n=1
1
3 n + 2n
3 n
= ∞
n=1
1
3
= 1/3
1 − 1/3 + 2/3
1 − 2/3 = 1 2 +2= 5 2
n n 2
+
3
[sum of two convergent geometric series]
27.
∞
n=1
n√
2=2+
√
2+
3 √ 2+ 4√ 2+··· diverges by the Test for Divergence since
29.
31.
33.
lim a n = lim
n√
2 = lim
n→∞ n→∞ n→∞ 21/n =2 0 =16= 0.
∞
n=1
n 2 +1
ln
diverges by the Test for Divergence since
2n 2 +1
n 2
lim an = lim ln +1
n→∞ n→∞ 2n 2 +1
∞
n=1
∞
n=1
=ln lim
n→∞
n 2 +1
=ln 1
2n 2 2
+1
6=0.
arctan n diverges by the Test for Divergence since lim
n→∞ a n = lim
n→∞ arctan n = π 2 6=0.
1
e = ∞ n 1
is a geometric series with first term a = 1 n n=1 e
e and ratio r = 1 e .Since|r| = 1 < 1, the series converges
e
1/e
to
1 − 1/e = 1/e
1 − 1/e · e
e = 1
e − 1 .ByExample6, ∞
n=1
∞ 1
n=1 e + 1
= ∞ 1
n n(n +1) n=1 e + ∞
n n=1
1
=1. Thus, by Theorem 8(ii),
n(n +1)
1
n(n +1) = 1
e − 1 +1= 1
e − 1 + e − 1
e − 1 =
e
e − 1 .