## Thomas' Calculus 13th Edition

converges to $\dfrac{1}{1-\ln x}$ for $|\ln x| \lt 1$ or, $e^{-1} \lt x \lt e$
Formula to calculate the sum of a geometric series is: $S=\dfrac{a}{1-r}$; Here, $a=1$ and common ratio $r =\ln x$ $S=\dfrac{a}{1-r}=\dfrac{1}{(1-\ln x)}$ Hence, the series converges to $\dfrac{1}{(1-\ln x)}$ for $|\ln x| \lt 1$ or, $e^{-1} \lt x \lt e$