#### Answer

$$\frac{1}{1-e}$$

#### Work Step by Step

Given
$$ \sum_{n=1}^{n=\infty} \left(e\right)^{-n}= \sum_{n=1}^{n=\infty} \left(\frac{1}{e}\right)^{-n}$$
Since the series is a geometric series with $|r|=\left|\frac{1}{e}\right|<1$, then the series converges and has the sum
\begin{align*}
S&= \frac{a_1}{1-r}\\
&=\frac{\frac{1}{e}}{1-\frac{1}{e}}\\
&=\frac{1}{1-e}
\end{align*}