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: 44


$\frac{x-sinx}{x^{3}}=\frac{1}{6}-\frac{1}{120}x^{2} +\frac{1}{5040}x^{4}-...=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$, for all values of $x$ $R=\infty$

Work Step by Step

$\frac{x-sinx}{x^{3}}=\frac{1}{x^{2}}-\frac{1}{x^{3}} \cdot sinx$ $=\frac{1}{x^{2}}-\frac{1}{x^{3}} \Sigma_{n=0}^{\infty}(-1)^{n}\frac{{(x)}^{2n+1}}{(2n+1)!}$ $=\frac{1}{x^{2}}+\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n+1)-3}}{(2n+1)!}$ $=\frac{1}{x^{2}}+\Sigma_{n=0}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$ $=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{{(x)}^{(2n-2)}}{(2n+1)!}$ $R=\infty$
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