Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.8 - Power Series - 11.8 Exercises - Page 751: 3

Answer

$R=1$ ; interval of convergence is $[-1,1)$

Work Step by Step

Let $a_{n}=(-1)^{n}nx^{n}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(-1)^{n+1}(n+1)x^{n+1}}{(-1)^{n}nx^{n}}|$ $=|x|$ converges if $|x|\lt 1$, so $R=1$ and the interval of convergence is from $-1$ to $+1$ and including $-1$ since it is the alternating harmonic series which converges. Hence, $R=1$ ; interval of convergence is $[-1,1)$
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