How to verify the equation $\frac{1}{e^{2\pi x} - 1} = \sum_{k = 1}^{\infty} e^{-2\pi kx}$

Question

May somebody tell me, why $$ \frac{1}{e^{2 \pi x} - 1} = \sum\limits_{k=1}^{\infty} e^{-2 \pi k x}? $$ It's obviously not Taylor's and Fourier's series, so how to prove it?

Answer 1

Geometric series with common ratio $e^{-2\pi x}$. See Wikipedia for the formula of such a sum.

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Note that your sum start at $k=1$, so we cannot exactly use the "famous" formula $\frac{1}{1-r}$, but rather $\frac{a}{1-r}$.

Answer 2

If $\;x>0\;$ , then $\;|e^{-2\pi x}|<1\;$ , and then we get:

$$\frac1{e^{2\pi x}-1}=\frac1{e^{2\pi x}}\frac1{1-e^{-2\pi x}}=\frac1{e^{2\pi x}}\sum_{k=0}^\infty e^{-2k\pi x}=\sum_{k=0}^\infty e^{-2\pi x(k+1)}=\sum_{k=1}^\infty e^{-2k\pi x}$$