Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.9 - Convergence of Taylor Series - Exercises 10.9 - Page 625: 27


$\Sigma_{n=1}^\infty \dfrac{ (-1)^{n-1}x^{(2n+1)}}{3n}$

Work Step by Step

Here, $ \dfrac{x^3}{3} \cdot \ln (1+x^2)=\dfrac{x^3}{3}[x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-...]$ This implies that $\dfrac{x^3}{3}[x^2-\dfrac{(x^2)^2}{2!}+\dfrac{(x^2)^3}{3!}-...]=\dfrac{x^3}{3}-\dfrac{x^5}{6}+\dfrac{x^7}{9}-...$ Hence, $\dfrac{x^3}{3} \cdot \ln (1+x^2)=\Sigma_{n=1}^\infty \dfrac{ (-1)^{n-1}x^{(2n+1)}}{3n}$
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