Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 785: 30

Answer

$$cos\frac{\sqrt \pi}{3}$$

Work Step by Step

Given: $$\Sigma_{n=0}^{\infty}\frac{(-1)^{n}\pi^{n}}{3^{2n}(2n)!}$$ As we know, $$cosx=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}$$ Re-write the given equation as $$\Sigma_{n=0}^{\infty}\frac{(-1)^{n}\pi^{n}}{3^{2n}(2n)!}=\Sigma_{n=0}^{\infty}\frac{(-1)^{n}(\sqrt \pi/3)^{2n}}{(2n)!}$$ Now we can see that $x=\frac{\sqrt \pi}{3}$, thus, the series sum is $$\Sigma_{n=0}^{\infty}\frac{(-1)^{n}\pi^{n}}{3^{2n}(2n)!}=cos\frac{\sqrt \pi}{3}$$
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