## Thomas' Calculus 13th Edition

Interval of Convergence : $e^{-1} \lt x \lt e$ and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$
Remember that the series will converge when $|f(x)|=|\ln x| \lt 1$ This implies that $-1 \lt \ln x \lt 1$ and $e^{-1} \lt x \lt e$ This means that the interval of convergence is: $e^{-1} \lt x \lt e$ Next, we will find the sum of the given geometric series $S=\dfrac{a}{1-r}=\dfrac{1}{1-\ln x}$ and $\Sigma_{n=0}^{\infty} (\ln x)^n=\dfrac{1}{1-\ln x}$