Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.2 SERIES ¤ 463
35. Using partial fractions, the partial sums of the series ∞
s n = n
=
i=2
1 − 1 3
n=2
2
n 2 − 1 are
2
(i − 1)(i +1) = n 1
i=2 i − 1 − 1
i +1
1
+
2 − 1 1
+
4
3 − 1
+ ···+
5
This sum is a telescoping series and s n =1+ 1 2 − 1
n − 1 − 1 n .
∞ 2
Thus,
n 2 − 1 = lim sn = lim 1+ 1
n→∞ n→∞ 2 − 1
n − 1 − 1
= 3 n 2 .
n=2
37. For the series ∞
n=1
3
n(n +3) , sn = n 3
i(i +3) = n 1
i − 1
i=1
1 −
1
4
+
1
2 − 1 5
+
1
3 − 1 6
+
1
4 − 1 7
Thus,
∞
n=1
39. For the series ∞
i=1
+ ···+
1
n−3 − 1 n
i +3
+
1
n − 3 − 1 1
+
n − 1 n − 2 − 1
n
[using partial fractions]. The latter sum is
1
− 1
n −2 n +1
+
1
− 1
n−1 n+2
+
1
− 1
n n+3
=1+ 1 + 1 − 1 2 3 n +1 n +2 n+3
[telescoping series]
3
n(n +3) = lim n = lim 1+ 1 + 1 − 1 =1+ 1 + 1 = 11 .
n→∞ n→∞
2 3 n +1 n +2 n+3
2 3 6 Converges
n=1
e 1/n − e 1/(n+1)
,
s n = n
e 1/i − e 1/(i+1)
i=1
=(e 1 − e 1/2 )+(e 1/2 − e 1/3 )+···+
e 1/n − e 1/(n+1) = e − e 1/(n+1)
[telescoping series]
∞
Thus,
e 1/n − e 1/(n+1) = lim s n = lim
e − e 1/(n+1) = e − e 0 = e − 1. Converges
n→∞ n→∞
n=1
41. 0.2 = 2 10 + 2
2
+ ··· is a geometric series with a =
102 10 and r = 1 10 .Itconvergesto a
1 − r = 2/10
1 − 1/10 = 2 9 .
43. 3.417 = 3 + 417
10 + 417
3 10 + ···.Now417
6 10 + 417
417
+ ··· is a geometric series with a = 3 106 10 and r = 1
3 10 . 3
It converges to
a
1 − r = 417/103
1 − 1/10 3 = 417/103
999/10 = 417
3 999
.Thus,3.417 = 3 +
417
999 = 3414
999 = 1138
333 .
45. 1.5342 = 1.53 + 42
10 + 42
42
+ ···.Now 4 106 10 + 42
42
+ ··· is a geometric series with a = 4 106 10 and r = 1
4 10 . 2
47.
It converges to
a
1 − r = 42/104
1 − 1/10 2 = 42/104
99/10 2 = 42
9900 .
Thus, 1.5342 = 1.53 + 42
9900 = 153
100 + 42
9900 = 15,147
9900 + 42
9900 = 15,189
9900
∞
n=1
or
5063
3300 .
x n
3 = ∞ x
n x
|x|
is a geometric series with r = , so the series converges ⇔ |r| < 1 ⇔ < 1 ⇔ |x| < 3;
n=1 n 3
3 3
that is, −3