## Calculus 8th Edition

The series converges when $x \lt 0$ and the sum is $\frac{1}{1-e^{x}}$.
Given: $\Sigma^{\infty}_{n=0}e^{nx}$ Here, $a=1$ and $r=e^{x}$ The series converges when $|r| \lt 1$ $|e^{x}| \lt 1$ $e^{x} \lt 1$ $x \lt ln_{e}1$ $x \lt 0$ Hence the series converges if $x \lt 0$ Sum can be calculated as follow: $\Sigma^{\infty}_{n=0}e^{nx} = \frac{a}{1-r}$ $=\frac{1}{1-e^{x}}$