Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.1 SEQUENCES ¤ 457
49. From the graph, it appears that the sequence converges to 1 2 .
As n →∞,
3+2n
2
a n =
8n 2 + n = 3/n 2 +2
8+1/n
so lim
n→∞ a n = 1 2 .
⇒
0+2
8+0 =
1
4 = 1 2 ,
n 2 cos n
51. From the graph, it appears that the sequence {a n } =
is
1+n 2
divergent, since it oscillates between 1 and −1 (approximately). To
prove this, suppose that {a n} converges to L. Ifb n =
n2
1+n 2 ,then
a n
{b n} converges to 1,and lim = L
n→∞ b n 1 = L. Buta n
=cosn,so
b n
a n
lim does not exist. This contradiction shows that {a n } diverges.
n→∞ b n
53. From the graph, it appears that the sequence approaches 0.
0 1.
57. If |r| ≥ 1, then{r n } diverges by (9), so {nr n } diverges also, since |nr n | = n |r n | ≥ |r n |.If|r| < 1 then
x
lim
x→∞ xrx = lim
x→∞ r −x
whenever |r| < 1.
H = lim
x→∞
1
= lim
(− ln r) r−x x→∞
r x
=0,so lim
− ln r n→∞ nrn =0, and hence {nr n } converges
59. Since {a n } is a decreasing sequence, a n >a n+1 for all n ≥ 1. Because all of its terms lie between 5 and 8, {a n } is a
bounded sequence. By the Monotonic Sequence Theorem, {a n} is convergent; that is, {a n} has a limit L. L must be less than
8 since {a n } is decreasing, so 5 ≤ L