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