Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Review - Exercises - Page 786: 53

Answer

$16$

Work Step by Step

$\frac{1}{\sqrt[4] {16-x}}=\frac{1}{2}\Sigma_{n=1}^\infty\frac{1.5.9....(4n-3)x^{n}}{2^{6n+1}n!}$ $$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{1.5.9....(4n-3)x^{n+1}(4n-3+4)}{2^{6n+1}n!}}{\frac{1.5.9....(4n-3)x^{n}}{2^{6n+1}n!}}|$$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(4n-1)x}{2^{6}(n+1)}|$ $=\lim\limits_{n \to \infty}|\frac{4x-x/n}{2^{6}+2^{6}/n)}|$ $=|\frac{4x}{2^{6}}|$ $=|\frac{x}{2^{4}}|$ $=|\frac{x}{2^{4}}|\lt 1$ $=|x|\lt 16$ Thus, the series converges when $|x|\lt 16$ and the radius of convergence is $16$.
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