Solução_Calculo_Stewart_6e

F.
456 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
25. a n = (−1)n−1 n
n 2 +1
Theorem 6.
= (−1)n−1
n +1/n ,so0 ≤ |an| = 1
n +1/n ≤ 1 → 0 as n →∞,soan → 0 by the Squeeze Theorem and
n
Converges
27. a n =cos(n/2). This sequence diverges since the terms don’t approach any particular real number as n →∞.
The terms take on values between −1 and 1.
29. a n =
(2n − 1)!
(2n +1)! = (2n − 1)!
(2n +1)(2n)(2n − 1)! = 1
→ 0 as n →∞. Converges
(2n +1)(2n)
31. a n = en + e −n
e 2n − 1 · e−n
e
e n − e −n → 0 as n →∞because 1+e−2n → 1 and e n − e −n →∞.
−n
=
1+e−2n
Converges
33. a n = n 2 e −n = n2
x 2
.Since lim
en x→∞ e x
= H 2x
lim
x→∞ e x
= H 2
lim =0, it follows from Theorem 3 that lim an =0.
x→∞ e Converges
x n→∞
35. 0 ≤ cos2 n
≤ 1
[since 0 ≤ cos 2 1 cos 2
n ≤ 1], so since lim
2 n 2 n n→∞ 2 =0, n
converges to 0 by the Squeeze Theorem.
n 2 n
37. a n = n sin(1/n) = sin(1/n)
sin(1/x) sin t
. Since lim = lim
1/n x→∞ 1/x t→0 + t
that {a n } converges to 1.
39. y =
1+ 2 x
⇒ ln y = x ln 1+ 2
,so
x x
1
− 2 x 2
ln(1 + 2/x) H 1+2/x
lim ln y = lim
= lim
x→∞ x→∞ 1/x x→∞ −1/x 2
lim 1+ 2 x
= lim
x→∞ x x→∞ eln y = e 2 ,sobyTheorem3, lim
n→∞
41. a n =ln(2n 2 +1)− ln(n 2 +1)=ln
= lim
x→∞
[where t =1/x] =1, it follows from Theorem 3
2
1+2/x =2
⇒
1+ 2 n
n
= e 2 . Convergent
2n 2 +1 2+1/n
2
=ln
→ ln 2 as n →∞.
n 2 +1 1+1/n 2
Convergent
43. {0, 1, 0, 0, 1, 0, 0, 0, 1,...} diverges since the sequence takes on only two values, 0 and 1, and never stays arbitrarily close to
either one (or any other value) for n sufficiently large.
45. a n = n!
2 = 1 n 2 · 2
2 · 3 (n − 1)
····· · n
2 2 2 ≥ 1 2 · n
2
[for n>1] = n 4
→∞as n →∞,so{an} diverges.
47. From the graph, it appears that the sequence converges to 1.
{(−2/e) n } converges to 0 by (9), and hence {1+(−2/e) n }
converges to 1+0=1.