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: 13


$$R=2$$ and Interval of convergence: $$[-2,2)$$

Work Step by Step

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