Solução_Calculo_Stewart_6e

F.
458 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
63. The terms of a n = n(−1) n alternate in sign, so the sequence is not monotonic. The first five terms are −1, 2, −3, 4,and−5.
Since lim |a n| = lim n = ∞, the sequence is not bounded.
n→∞ n→∞
65. a n = n defines a decreasing sequence since for f(x) =
x
x 2
n 2 +1 x 2 +1 , f 0 +1 (1) − x(2x)
(x) =
(x 2 +1) 2 = 1 − x2
(x 2 +1) 2 ≤ 0
for x ≥ 1. The sequence is bounded since 0 a n and 0 a n ⇒ 1
a n+1
< 1
a n
⇒
− 1
a n+1
> − 1
a n
.Nowa n+2 =3− 1
a n+1
> 3 − 1
a n
= a n+1 ⇔ P n+1 . This proves that {a n } is increasing and bounded
above by 3,so1=a 1 1,soL = 3+√ 5
2
.
71. (a) Let a n be the number of rabbit pairs in the nth month. Clearly a 1 =1=a 2 .Inthenth month, each pair that is
2 or more months old (that is, a n−2 pairs) will produce a new pair to add to the a n−1 pairs already present. Thus,
a n = a n−1 + a n−2 ,sothat{a n } = {f n }, the Fibonacci sequence.
(b) a n = fn+1
⇒ a n−1 = fn fn−1 + fn−2
= =1+ fn−2
1
=1+
=1+ 1 .IfL = lim
f n f n−1 f n−1
f n−1 f n−1 /f n−2 a an,
n−2 n→∞
then L = lim a n−1 and L = lim a n−2,soL must satisfy L =1+ 1
n→∞ n→∞ L ⇒ L2 − L − 1=0 ⇒ L = 1+√ 5
2
[since L must be positive].
n
5
73. (a) From the graph, it appears that the sequence
n!
n 5
converges to 0,thatis, lim
n→∞ n! =0.