Answer
Converges to $\dfrac{1}{1-\ln x}$ for $|\ln x| \lt 1$ or, $e^{-1} \lt x \lt e$
Work Step by Step
The sum of a geometric series can be found as:
$S=\dfrac{a}{1-r}$
We have a convergent geometric series with first term, $a=1$ and common ratio $r =\ln x$
$S=\dfrac{1}{1-\ln x}$
Hence, the given series converges to $\dfrac{1}{1-\ln x}$ for $|\ln x| \lt 1$ or, $e^{-1} \lt x \lt e$
