Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.8 Power Series - 11.8 Exercises - Page 791: 17


$R=2$ ; interval of convergence is $[-4,0)$

Work Step by Step

Let $a_{n}=\frac{(x+2)^{n}}{2^{n}lnn}$, then $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{(x+2)^{n+1}}{2^{n+1}ln(n+1)}}{\frac{(x+2)^{n}}{2^{n}lnn}}|$ $=\frac{|x+2|}{2}$ Hence, $R=2$ ; interval of convergence is $[-4,0)$
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