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: 16


$R=2$ and Interval of convergence: $(-1,3]$

Work Step by Step

Given: $$a_n=\frac{(x-1)^{n}}{(2n-1)2^{n}}$$$$|\frac{a_{n+1}}{a_{n}}|=|\dfrac{\dfrac{(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\dfrac{(x-1)^{n}}{(2n-1)2^{n}}}|$$ Take limits on both sides, we have $$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(x-1)^{n+1}}{(2(n+1)-1)2^{n+1}}}{\dfrac{(x-1)^{n}}{(2n-1)2^{n}}}|=\frac{x-1}{2}$$ Hence, $R=2$ and Interval of convergence: $(-1,3]$
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