Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.5 ALTERNATING SERIES ¤ 473
45. Yes. Since a n is a convergent series with positive terms, lim
n→∞ an =0by Theorem 11.2.6, and b n = sin(a n) is a
b n sin(a n)
series with positive terms (for large enough n). We have lim = lim =1> 0 by Theorem 3.3.2. Thus, b n
n→∞ a n n→∞ a n
is also convergent by the Limit Comparison Test.
11.5 Alternating Series
1. (a) An alternating series is a series whose terms are alternately positive and negative.
(b) An alternating series ∞ (−1) n−1 b n converges if 0 0, {b n} is decreasing, and lim b n =0,sothe
n→∞
5.
7.
n=1
series converges by the Alternating Series Test.
∞
a n = ∞
n=1
n=1
(−1) n−1 1
2n +1 = ∞
n=1
series converges by the Alternating Series Test.
∞
a n = ∞ (−1) n 3n − 1
2n +1 = ∞ (−1) n b n.Now lim
n=1
n=1
n=1
(−1) n−1 b n .Nowb n = 1
2n +1 > 0, {b n} is decreasing, and lim
n→∞ b n =0,sothe
n→∞
(in fact the limit does not exist), the series diverges by the Test for Divergence.
3 − 1/n
bn = lim
n→∞ 2+1/n = 3 6=0.Since lim an 6= 0
2 n→∞
9. b n = n
10 > 0 for n ≥ 1. {b n} is decreasing for n ≥ 1 since
n
x
0 10 x (1) − x · 10 x ln 10
= = 10x (1 − x ln 10) 1 − x ln 10
= < 0 for 1 − x ln 10 < 0 ⇒ x ln 10 > 1 ⇒
10 x (10 x ) 2 (10 x ) 2 10 x
x> 1 ≈ 0.4. Also, lim bn = lim
ln 10 n→∞ n→∞
converges by the Alternating Series Test.
n
10 n = lim
x→∞
x
10 x H = lim
x→∞
x
10 x ln 10 =0. Thus, the series ∞
n=1
(−1) n
n
10 n
11. b n = n2
> 0 for n ≥ 1.
n 3 +4
x
2
13.
x 3 +4
lim
n→∞ b n = lim
n→∞
∞
(−1) n n
n=2
{bn} is decreasing for n ≥ 2 since
0
= (x3 + 4)(2x) − x 2 (3x 2 )
(x 3 +4) 2 = x(2x3 +8− 3x 3 )
ln n . lim
n→∞
= x(8 − x3 )
< 0 for x>2. Also,
(x 3 +4) 2 (x 3 +4)
2
1/n
1+4/n =0.Thus,theseries ∞
(−1) n+1 n 2
converges by the Alternating Series Test.
3 n 3 +4
n
ln n = lim
x→∞
x
ln x
H
= lim
x→∞
n=1
1
= ∞, so the series diverges by the Test for Divergence.
1/x