Solução_Calculo_Stewart_6e

F.
484 ¤ CHAPTER 11 INFINITE SEQUENCES AND SERIES
TX.10
37. s 2n−1 =1+2x + x 2 +2x 3 + x 4 +2x 5 + ···+ x 2n−2 +2x 2n−1
=1(1+2x)+x 2 (1 + 2x)+x 4 (1 + 2x)+···+ x 2n−2 (1 + 2x) =(1+2x)(1 + x 2 + x 4 + ···+ x 2n−2 )
=(1+2x) 1 − x2n
[by (11.2.3)] with r = x 2 ] → 1+2x as n →∞ [by (11.2.4)], when |x| < 1.
1 − x 2 1 − x2 Also s 2n = s 2n−1 + x 2n → 1+2x
1 − x since 2 x2n → 0 for |x| < 1. Therefore, s n → 1+2x since s2n and s2n−1 both
1 − x2 approach 1+2x as n →∞. Thus, the interval of convergence is (−1, 1) and f(x) =1+2x
1 − x2 1 − x . 2
39. We use the Root Test on the series
c n x n n
.Weneed lim |cn x n n
| = |x| lim |cn | = c |x| < 1 for convergence, or
n→∞
n→∞
|x| < 1/c,soR =1/c.
41. For 2