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: 39

Answer

Radius of convergence is $8$

Work Step by Step

We need to apply the Ratio Test to the series. $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{\dfrac{(n+1)a_nx}{4(2n+1}}{a_n}$ or, $=|x| \lim\limits_{n \to \infty} \dfrac{n+1}{8n+4}$ or, $=|x| \lim\limits_{n \to \infty} \dfrac{1+1/n}{8+4/n}$ or, $=\dfrac{1}{8}|x|$ Now, $\dfrac{1}{8}|x| \lt 1 \implies |x| \lt 8$ Thus, the radius of convergence is $8$.
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