Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 736: 63

The series converges when $x \lt 0$ and the sum is $\frac{1}{1-e^{x}}$.

$\Sigma^{\infty}_{n=0}e^{nx}$ $a=1$ 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 $x \lt 0$ $\Sigma^{\infty}_{n=0}e^{nx} = \frac{a}{1-r}$ $=\frac{1}{1-e^{x}}$