Solução_Calculo_Stewart_6e

F.
472 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES TX.10
1
31. Use the Limit Comparison Test with a n =sin and b n = 1 n n .Then a n and b n are series with positive terms and
33.
35.
a n
lim
n→∞ b n
sin(1/n) sin θ
= lim =lim =1> 0. Since ∞ b n is the divergent harmonic series,
n→∞ 1/n θ→0 θ
n=1
∞
sin (1/n) also diverges.
n=1
[Note that we could also use l’Hospital’s Rule to evaluate the limit:
−1/x
sin(1/x) H cos(1/x) ·
2
lim = lim
= lim
x→∞ 1/x x→∞ −1/x cos 1 2
x→∞ x =cos0=1.]
10
n=1
1
√
n4 +1 = 1 √
2
+ 1 √
17
+ 1 √
82
+ ···+
R 10 ≤ T 10 ≤
10
n=1
∞
10
1
√ 10,001
≈ 1.24856. Now
1
dx = lim − 1 t
= lim
− 1 x2 t→∞ x t→∞
10
t + 1
= 1
10 10 =0.1.
1
1+2 = 1 n 3 + 1 5 + 1 9 + ···+ 1
1025 ≈ 0.76352. Now 1
1+2 < 1 , so the error is
n 2n R 10 ≤ T 10 =
∞
n=11
1
2 n = 1/211
1 − 1/2
37. Since d n
10 n ≤ 9
10 n for each n,andsince ∞
[geometric series] ≈ 0.00098.
n=1
will always converge by the Comparison Test.
1
√
n4 +1 < 1 √
n
4 = 1 n 2 ,sotheerroris
9
10 n is a convergent geometric series |r| = 1
10 < 1 , 0.d 1 d 2 d 3 ...= ∞
39. Since a n converges, lim an =0,sothereexistsN such that |an − 0| < 1 for all n>N ⇒ 0 ≤ an < 1 for
n→∞
all n>N ⇒ 0 ≤ a 2 n ≤ a n . Since a n converges, so does a 2 n by the Comparison Test.
a n
41. (a) Since lim = ∞, there is an integer N such that an
> 1 whenever n>N.(TakeM =1in Definition 11.1.5.)
n→∞ b n
b n
Then a n >b n whenever n>Nand since b n is divergent, a n is also divergent by the Comparison Test.
(b) (i) If a n = 1
ln n and b n = 1 a n
for n ≥ 2,then lim = lim
n n→∞ b n n→∞
so by part (a),
(ii) If a n = ln n
n
∞
n=2
1
is divergent.
ln n
and b n = 1 n ,then ∞
so ∞ a n diverges by part (a).
n=1
n=1
n
ln n = lim
x→∞
x
ln x
H
= lim
x→∞
n=1
1
1/x = lim
x→∞ x = ∞,
d n
10 n
a n
b n is the divergent harmonic series and lim = lim ln n = lim ln x = ∞,
n→∞ b n n→∞ x→∞
a n
43. lim nan = lim
n→∞ n→∞ 1/n , so we apply the Limit Comparison Test with bn = 1 .Since lim nan > 0 we know that either both
n n→∞
series converge or both series diverge, and we also know that
divergent.
∞
n=1
1
n diverges [p-series with p =1]. Therefore, a n must be