Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 31


$1-\frac{1}{4}x-\Sigma_{n=2}^{\infty}\frac{3.7.......(4n-5)x^{n}}{4^{n}n!}$ and $R=1$

Work Step by Step

$\sqrt[4] {1-x}=(1-x)^{1/4}=(1+(-x))^{1/4}$ $=1-\frac{1}{4}x-\Sigma_{n=2}^{\infty}\frac{3.7.......(4n-5)x^{n}}{4^{n}n!}$ $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\dfrac{3.7.......(4n-5)(4(n+1)-5)x^{n}}{4^{n+1}(n+1)!}}{\dfrac{3.7.......(4n-5)x^{n}}{4^{n}n!}}|$ $=\lim\limits_{n \to\infty}|\frac{(4n-1)x}{4(n+1)}|$ $=|x|$ The series will converge when $|x|\lt 1$, so $R=1$.
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