Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 11.8 POWER SERIES ¤ 483
x n
27. If a n =
1 · 3 · 5 · ··· ·(2n − 1) ,then
lim
n→∞ a n+1 = lim
x n+1
n→∞ 1 · 3 · 5 ·····(2n − 1)(2n +1)
a n
Ratio Test, the series ∞
n=1
1 · 3 · 5 ·····(2n − 1)
· = lim
x n
n→∞
|x|
=0< 1. Thus, by the
2n +1
x n
converges for all real x and we have R = ∞ and I =(−∞, ∞).
1 · 3 · 5 ·····(2n − 1)
29. (a) We are given that the power series ∞
n=0 c nx n is convergent for x =4. So by Theorem 3, it must converge for at least
−4