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: 21


$ln(2)+\Sigma_{n=1}^{\infty}\frac{(x-2)^{n}}{2^{n}n}(-1)^{n+1}$ $R=2$

Work Step by Step

$lnx=ln(2)+\Sigma_{n=1}^{\infty}\frac{(x-2)^{n}}{2^{n}n}(-1)^{n+1}$ $\lim\limits_{n \to \infty}|\dfrac{a_{n+1}}{a_{n}}=\lim\limits_{n \to \infty}|\frac{\frac{(x-2)^{n+1}(-1)^{n+2}}{2^{n+1}(n+1)}}{\frac{(x-2)^{n}(-1)^{n+1}}{2^{n}n}}|$ $=\lim\limits_{n \to\infty}|\frac{x-2}{2}|$ The series will converge when $|\frac{x-2}{2}|\lt 1$, or $|x-2|\lt 2$ so $R=2$.
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