University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 9 - Section 9.7 - Power Series - Exercises - Page 531: 38


Radius of convergence is $\dfrac{9}{4}$

Work Step by Step

We need to apply the Ratio Test to the series. $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{(2n+2)^2x}{(3n+2)^2}$ or, $=|x| \lim\limits_{n \to \infty}(\dfrac{2n+2}{3n+2})^2$ or, $=|x| \lim\limits_{n \to \infty}(\dfrac{2+2/n}{3+2/n})^2$ or, $=\dfrac{4}{9}|x|$ Now, $\dfrac{4}{9}|x|| \lt 1 \implies |x| \lt \dfrac{9}{4}$ Thus, the radius of convergence is $\dfrac{9}{4}$
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