Calculus: Early Transcendentals 8th Edition

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