Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.9 - Representations of Functions as Power Series - 11.9 Exercises - Page 757: 16

Answer

$\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{6n+5}}{2n+1}$, $R=1$

Work Step by Step

$f(x)=x^{2}tan^{-1}(x^{3})=\sum_{n=0}^{\infty}(-1)^{n}\frac{x^{6n+5}}{2n+1}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{6(n+1)+5}}{2(n+1)+1}}{\frac{x^{6n+5}}{2n+1}}|$ $=|x^{6}|\lt 1$ The given series converges with $R=1$
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