Answer
Interval of Convergence: $e^{-1} \lt x \lt e$
and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$
Work Step by Step
The series will converge when
$|f(x)|=|\ln x| \lt 1$
So, $-1 \lt \ln x \lt 1$
$ \implies e^{-1} \lt x \lt e$
$ \implies e^{-1} \lt x \lt e$
Now, the sum of the given geometric series is:
$S=\dfrac{a}{1-r}=\dfrac{1}{1-\ln x}$
Thus, the Interval of Convergence is:
$e^{-1} \lt x \lt e$
and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$
