## Thomas' Calculus 13th Edition

$\dfrac{e}{(e-1)}$
The sum of the geometric series is given by $s_n=\dfrac{a}{(1-r)}$ Here, $a$ is initial term and $r$ is common ratio. A geometric series is to be convergent when $|r|\lt 1$ and divergent when $|r|\gt 1$. From the given problem, we have $r=e^{-1} \lt 1$ Thus, $s_n=\dfrac{a}{(1-r)}=\dfrac{1}{(1-e^{-1})}$ Hence, $s_n=\dfrac{e}{(e-1)}$