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: 74


$\frac{\sqrt 3}{2}$

Work Step by Step

Given: $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}$ As we know $cosx=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n}}{2n!}$ Thus, $cos(\frac{\pi}{6})=\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}$ or $\Sigma_{n=0}^{\infty}{(-1)^{n}}\dfrac{\pi^{2n}}{6^{2n}(2n)!}=\frac{\sqrt 3}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.