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: 80



Work Step by Step

Given: $\frac{1}{1.2}-\frac{1}{3.2^{3}}+\frac{1}{5.2^{5}}+....=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{1}{(2n+1).2^{2n+1}}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\dfrac{\frac{1}{2^{2n+1}}}{(2n+1)}$ Since, $tan^{-1}x=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)}$ Thus, $tan^{-1}(\frac{1}{2})=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(1/2)^{2n+1}}{(2n+1)}$
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