Prove geometric sum identity

Question

Prove that ${ \sum_{n=1}^{\infty}{x e^{-nx} }} = { \frac{x}{ e^{x} - 1 } }$ for $x \in [0,1]$.

I know that with geometric sums ${ \sum_{n=0}^{\infty}{x^n }} = { \frac{1}{1- x } } $but which manipulations should I do on the known sum, in order to prove the identity above?

Thanks.

Answer 1

$$\sum_{n=1}^{\infty}x\,e^{-nx}=x\sum_{n=1}^{\infty}\,\left(e^{-x}\right)^n=x\frac{e^{-x}}{1-e^{-x}}=\frac{x}{e^x-1}$$