Solução_Calculo_Stewart_6e

F.
TX.10
11 INFINITE SEQUENCES AND SERIES
11.1 Sequences
1. (a) A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers.
(b) The terms a n approach 8 as n becomes large. In fact, we can make a n as close to 8 as we like by taking n sufficiently
large.
(c) The terms a n become large as n becomes large. In fact, we can make a n as large as we like by taking n sufficiently large.
3. a n =1− (0.2) n , so the sequence is {0.8, 0.96, 0.992, 0.9984, 0.99968,...}.
5. a n = 3(−1)n
−3
, so the sequence is
n!
1 , 3 2 , −3
6 , 3 24 , −3
120 ,... =
−3, 3 2 , −1 2 , 1 8 , − 1
40 ,... .
7. a 1 =3, a n+1 =2a n − 1. Eachtermisdefined in terms of the preceding term.
9.
a 2 =2a 1 − 1=2(3)− 1=5. a 3 =2a 2 − 1 = 2(5) − 1=9. a 4 =2a 3 − 1=2(9)− 1=17.
a 5 =2a 4 − 1=2(17)− 1=33. The sequence is {3, 5, 9, 17, 33,...}.
1,
1
3 , 1 5 , 1 7 , 1 9 ,... . The denominator of the nth term is the nth positive odd integer, so a n = 1
2n − 1 .
11. {2, 7, 12, 17,...}. Each term is larger than the preceding one by 5,soa n = a 1 + d(n − 1) = 2 + 5(n − 1) = 5n − 3.
13.
1, −
2
3 , 4 9 , − 8 27 ,... . Eachtermis− 2 3 times the preceding one, so a n = − 2 3
n−1
.
15. The first six terms of a n = n
2n +1 are 1 3 , 2 5 , 3 7 , 4 9 , 5
11 , 6
13 . It appears that the sequence is approaching 1 2 .
lim
n→∞
n
2n +1 = lim
n→∞
1
2+1/n = 1 2
17. a n =1− (0.2) n ,so lim
n→∞ a n =1− 0=1by (9).
Converges
19. a n = 3+5n2
n + n 2 = (3 + 5n2 )/n 2
(n + n 2 )/n = 5+3/n2
2 1+1/n
5+0
,soan → =5as n →∞. Converges
1+0
21. Because the natural exponential function is continuous at 0, Theorem 7 enables us to write
lim
n→∞ a n = lim
n→∞ e1/n = e lim n→∞(1/n) = e 0 =1.
Converges
23. If b n = 2nπ ,then lim bn = lim
1+8n n→∞ n→∞
Theorem 7, lim
n→∞ tan 2nπ
1+8n
=tan
(2nπ)/n
(1 + 8n)/n = lim
n→∞
lim
n→∞
2π
1/n +8 = 2π 8 = π 4 .Sincetan is continuous at π 4 ,by
2nπ
=tan π 1+8n 4 =1. Converges 455