Show that $\frac{1}{e^{t}-1} = \sum_{n=1}^{\infty} e^{-n t} $

Question

Show that $$ \frac{1}{e^{t}-1} = \sum_{n=1}^{\infty} e^{-n t} $$

I am a bit confused because we have a series similar to the geometric series.

What is the proof of this an equity?

Reference:

• http://www.math.harvard.edu/archive/213b_spring_05/functional_equation_by_residues.pdf

Answer 1

It is a geometric series with a ratio of $e^{-t}$:

$$\sum_{n=1}^{\infty}e^{-nt}=\lim_{n \to \infty}e^{-t}\frac{1-e^{-nt}}{1-e^{-t}}= e^{-t}\frac{1}{1-e^{-t}} = \frac{1}{e^{t}(1-e^{-t})} = \frac{1}{e^{t}-1}$$

Answer 2

For $t > 0$, $0 < e^{-t} < 1$. Thus, $$ \frac{1}{e^t - 1} = \frac{e^{-t}}{1 - e^{-t}} = e^{-t}\sum_{n=0}^{\infty}e^{-nt}=\sum_{n=0}^{\infty}e^{-(n+1)t} = \sum_{n=1}^{\infty}e^{-nt} $$