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: 27

Answer

$R=\infty$ ; interval of convergence is $(-\infty,\infty)$

Work Step by Step

Let $a_{n}=\frac{x^{n}}{1.3.5.....(2n-1)}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{n+1}}{1.3.5.....(2(n+1)-1)}}{\frac{x^{n}}{1.3.5.....(2n-1)}}|$ $=\lim\limits_{n \to \infty}\frac{|x|}{2n+1}\lt 1$ $=0\lt 1$ Hence, $R=\infty$ ; interval of convergence is $(-\infty,\infty)$
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