Answer
$=\frac{\sqrt 2}{2}$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}$
$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}=sin(\frac{\pi}{4})$
$=\frac{\sqrt 2}{2}$
Calculus: Early Transcendentals 8th Edition
by Stewart, James
Published by
Cengage Learning
ISBN 10:
1285741552
ISBN 13:
978-1-28574-155-0
Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 772: 77
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