How does $\displaystyle \frac{e^{-x}}{1-e^{-x}}$ become $\sum_{k=1}^{\infty}e^{-kx}$?

Question

How does $\displaystyle \frac{e^{-x}}{1-e^{-x}}$ become $\sum_{k=1}^{\infty}e^{-kx}$?

I know $\displaystyle \frac{e^{-x}}{1-e^{-x}}=\frac{1}{e^x-1}=\left(\sum_{k=0}^{\infty} \frac{x^k}{k!}-1\right)^{-1}=\left(\sum_{k=1}^{\infty} \frac{x^k}{k!}\right)^{-1}$ and we know $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!}=e^x$

Answer 1

Recall the geometric series (see Wikipedia): for any $y$ with $|y|<1$, $$\frac{1}{1-y}=1+y+y^2+\cdots=\sum_{k=0}^\infty y^k.$$ Therefore, for any such $y$, we also have $$\frac{y}{1-y}=y+y^2+y^3+\cdots=\sum_{k=1}^\infty y^k.$$ Now let $y=e^{-x}$ (though observe that we need $x>0$ to have $e^{-x}<1$).

Answer 2

Hint:

What is the series for $$\frac{1}{ x - 1}$$?

Answer 3

You have if $|a|<1$: $$ \sum_{i=0}^\infty a^i = \frac{1}{1-a} $$ therefore: $$ \frac{e^{-x}}{1-e^{-x}}= e^{-x}\sum_{i=0}^\infty \left({e^{-x}}\right)^i = \sum_{n=1}^\infty e^{-nx}. $$