Calculus: Early Transcendentals 8th Edition

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

Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 33


$\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}$ and $R=2$

Work Step by Step

$\frac{1}{(2+x)^{3}}=(2+x)^{-3}=2^{-3}(1+\frac{x}{2})^{-3}=\frac{1}{8}(1+\frac{x}{2})^{-3}$ $=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}$ $$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(n+2)(n+3)x^{n+1}}{2^{(n+5)}}}{\frac{(n+1)(n+2)x^{n}}{2^{(n+4)}}}|$$ $=\lim\limits_{n \to\infty}|\frac{(n+3)x}{2(n+2)}|$ $=|\frac{x}{2}|$ The series will converge when $|\frac{x}{2}|\lt 1$, or $|x|\lt 2$ so $R=2$.
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