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: 14

Answer

Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{2^{n}x^{n}}{n!}$ and $R=\infty$

Work Step by Step

$f(x)=e^{-2x}=\Sigma_{n=0}^{\infty}(-1)^{n}\frac{2^{n}x^{n}}{n!}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{2^{n+1}.x^{n+1}}{(n+1)!}}{\frac{2^{n}x^{n}}{n!!}}|$ $=\lim\limits_{n \to\infty}|(\frac{2x}{n+1})|$ $=0 \lt 1$ Maclaurin series is: $\Sigma_{n=0}^{\infty}(-1)^{n}\frac{2^{n}x^{n}}{n!}$ and $R=\infty$
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