Answer
By Th.9.6, the series converges.
Work Step by Step
$\displaystyle \sum_{n=1}^{\infty}e^{-n}=\sum_{n=1}^{\infty}(\frac{1}{e})^{n}$
is a geometric series with
$|r|=\displaystyle \frac{1}{e}\lt 1$.
By Th.9.6, the series converges.