How does $\displaystyle \frac{e^{-x}}{1-e^{-x}}$ become $\sum_{k=1}^{\infty}e^{-kx}$? 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 $\displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!}=e^x$

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 $e^{-x}<1$).
