Answer
$\frac{\sqrt 2}{2}$
Work Step by Step
Given: $\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}$
$\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}=sin(\frac{\pi}{4})$
$=\frac{\sqrt 2}{2}$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 11 - Infinite Sequences and Series - 11.10 Exercises - Page 790: 67
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