## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 11 - Section 11.10 - Taylor and Maclaurin Series - 11.10 Exercises - Page 771: 3

#### Answer

$f(x)=\Sigma_{n=0}^{\infty}(n+1)x^{n}$ and $R=1$

#### Work Step by Step

Given: $f^{n}(0)=(n+1)!$ for $n=0,1,2,...$ Maclaurin series: $f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}(x)^{2}+...$ $f(x)=1!+\frac{2!}{1!}x+\frac{3!}{2!}x^{2}+\frac{4!}{3!}x^{2}+...$ $=1+2x+3x^{2}+4x^{3}+...$ From this pattern we get $f(x)=\Sigma_{n=0}^{\infty}(n+1)x^{n}$ Use the ratio test: $\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\frac{(n+2)x^{n+1}}{(n+1)x^{n}}|$ $=\lim\limits_{n \to \infty}|\frac{(1+2/n)x}{(1+1/n)}|$ $=|x|$ This series converges when $|x| \lt 1$ Thus, the series has a radius of convergence of $1$ .

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