Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - Review - Exercises - Page 825: 34


$\frac{1}{e}\lt x\lt e$

Work Step by Step

A geometric series with common ratio $r$ converges only when $|r|\lt 1$.The sum of a geometric series $\Sigma _{n=1}^{\infty}ar^{n-1}$ equals $a/1-r$ The given series $\Sigma _{n=1}^{\infty}(ln(x))^{n}$ is a geometric series with common ratio $r=ln x$ Therefore, the series will converge when $|r|=lnx \lt 1$ This implies, $-1\lt lnx\lt 1$ $e^{-1}\lt e^{lnx}\lt e^{1}$ $\frac{1}{e}\lt x\lt e$ Hence, the series converges when $\frac{1}{e}\lt x\lt e$
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