University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.9 - Convergence of Taylor Series - Exercises - Page 542: 28


$(2) \Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{2n+1}$

Work Step by Step

We know that the Maclaurin Series for $ \ln (1+x)$ is defined as: $ \ln (1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...$ and $ \ln (1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-...$ Now, $ \ln (1+x)-\ln (1-x)=(x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-...)-(-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-...)$ Thus, $=2x+\dfrac{2x^3}{3}+\dfrac{2x^5}{5}+...$ or, $(2) \Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{2n+1}$
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