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: 34


$1-\frac{3}{4}x-3\Sigma_{n=2}^{\infty}\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}$ and $R=1$

Work Step by Step

${(1-x)^{3/4}}=(1+(-x))^{3/4}$ $=1-\frac{3}{4}x-3\Sigma_{n=2}^{\infty}\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}$ $$\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{5.9.13......(4n-7)(4(n+1)-7)x^{n+1}}{4^{n+1}(n+1)!}}{\frac{5.9.13......(4n-7)x^{n}}{4^{n}n!}}|$$ $=|x|$ The series will converge when $|x|\lt 1$ so $R=1$.
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