Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 11.10 TAYLOR AND MACLAURIN SERIES D 83
00 n.-1
and converges for x = - 1 (Alternating Series Test), so the interval of convergence is [-1, 1). !" (x) = I: ~ diverges
n=l n+ 1
I
at both 1 and - 1 (Test for Divergence) since lim ~ = 1 =/= 0, so its interval ofc~nvergence is ( - 1, 1).
n--.oo n+ 1
• 00 ·X2n+ l 1
41. By Example 7, tan- 1 x = L: ( -1)"-- for lxl < 1. In particular, for x = r.;• we
n=O 2n+ 1 y3
00
7r - 1 ( 1 ) n (1/ v'3) 2 n +l ~ ( )" (1)" 1 1
have 6 = tan y3 = n~O 1
( - ) 2n + 1 = n~O 1
- 3 y3 2n + 1' SO
7r=~f (-1)" =2v'3f (-1)" .
y'3 n = O (2n + 1)3" n=O (2n + 1)3"
11.10 Taylor and Maclaurin Series
oo J