Solução_Calculo_Stewart_6e

F.
506 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
29.
∞
[tan −1 (n +1)− tan −1 n]= lim
n=1
31. 1 − e + e2
2! − e3
3! + e4
4! − ···= ∞
n→∞ s n
= lim
n→∞ [(tan−1 2 − tan −1 1) + (tan −1 3 − tan −1 2) + ···+(tan −1 (n +1)− tan −1 n)]
= lim
n→∞ [tan−1 (n +1)− tan −1 1] = π 2 − π 4 = π 4
n=0
(−1) n e n
n! = ∞
n=0
(−e) n
n!
= e −e since e x = ∞
33. cosh x = 1 2 (ex + e −x )= 1 ∞
x n
2 n=0 n! + ∞ (−x) n
n=0 n!
= 1 1+x ···
+ x2
2
2! + x3
3! + x4
4! + +
1 ···
− x + x2
2! − x3
3! + x4
4! −
= 1 2+2· ···
x2 x4
+2·
2 2! 4! + =1+ 1
2 x2 + ∞ x 2n
n=2 (2n)! ≥ 1+ 1 2 x2 for all x
35.
37.
∞
n=1
(−1) n+1
n 5 =1− 1
32 + 1
243 − 1
1024 + 1
3125 − 1
7776 + 1
16,807 − 1
32,768 + ···.
Since b 8 = 1 8 5 = 1
32,768 < 0.000031, ∞
∞
n=1
1
2+5 ≈ 8
n
R 8 = ∞
n=9
n=1
1
2+5 < ∞
n
n=1
(−1) n+1
n 5 ≈ 7
n=1
(−1) n+1
n 5 ≈ 0.9721.
1
2+5 ≈ 0.18976224. To estimate the error, note that 1
n
n=9
39. Use the Limit Comparison Test. lim
n=0
x n
n!
for all x.
2+5 < 1 , so the remainder term is
n 5n 1
5 n = 1/59
1 − 1/5 =6.4 × 10−7 geometric series with a = 1
5 9 and r = 1 5
.
n→∞
n +1
n
a n
an
n +1
= lim = lim 1+ 1
=1> 0.
n→∞ n n→∞ n
Since |a n | is convergent, so is n +1
a n , by the Limit Comparison Test.
n
41. lim
n→∞
a n+1
a n
|x +2|
n+1
= lim
n→∞ (n +1)4 · n+1
n 4 n
|x +2| n = lim
n→∞
n
n +1
|x +2| < 4 ⇔ −4