Answer
The series converges when $x \lt 0$ and the sum is $\frac{1}{1-e^{x}}$.
Work Step by Step
$\Sigma^{\infty}_{n=0}e^{nx}$
$a=1$ and $r=e^{x}$
$|r| \lt 1$
$|e^{x}| \lt 1$
$e^{x} \lt 1$
$x \lt ln_{e}1$
$x \lt 0$
Hence the series converges if $x \lt 0$
$\Sigma^{\infty}_{n=0}e^{nx} = \frac{a}{1-r}$
$=\frac{1}{1-e^{x}}$