Calculus: Early Transcendentals 8th Edition

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


$=\frac{\sqrt 2}{2}$

Work Step by Step

Given: $\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}$ $\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\pi^{2n+1}}{4^{2n+1}(2n+1)!}=sin(\frac{\pi}{4})$ $=\frac{\sqrt 2}{2}$
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