Calculus 8th Edition

Published by Cengage

Chapter 11 - Infinite Sequences and Series - 11.10 Taylor and Maclaurin Series - 11.10 Exercises - Page 811: 17

Answer

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

Work Step by Step

$f(x)=sinhx=\Sigma_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$ $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{\frac{x^{2n+3}}{(2n+3)!}}{\frac{x^{2n+1}}{(2n+1)!}}|$ $=\lim\limits_{n \to\infty}|\frac{x^{2}}{(2n+2)(2n+3)}|$ $=\lim\limits_{n \to\infty}|\frac{x^{2}}{\infty}|$ $=0\lt 1$ Therefore, the Maclaurin's series converges for all values of $x$. Maclaurin's series is: $\Sigma_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$ and $R=\infty$

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