Answer
$\dfrac{e}{(e-1)}$
Work Step by Step
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, $s_n=\dfrac{e}{(e-1)}$
