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: 15


Maclaurin's series is: $\Sigma_{n=0}^{\infty}\frac{{(ln2})^{n}x^{n}}{(n)!}$ and $R=\infty$

Work Step by Step

$f(x)=2^{x}=\Sigma_{n=0}^{\infty}\frac{(ln2)^{n}x^{n}}{(n)!}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{(ln2)^{n+1}x^{n+1}}{(n+1)!}}{\frac{(ln2)^{n}x^{n}}{(n)!}}|$ $=\lim\limits_{n \to\infty}|\frac{(ln2).x}{n+1}|$ $=|x|\lt 1$ Therefore, the Maclaurin's series converges for all values of $x$. Maclaurin's series is: $\Sigma_{n=0}^{\infty}\frac{{(ln2})^{n}x^{n}}{(n)!}$ and $R=\infty$
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