I'm trying to prove the Lipschitz Summation Formula, that is for $k\in \mathbb{Z}$ and $z\in \mathbb{C}$ with $Re(z)>0$; $$\sum_{n\in \mathbb{Z}}\frac{1}{(z+n)^{k+1}}=\frac{(-2\pi i)^{k+1}}{k!}\sum_{n=1}^{\infty}n^{k}e^{2\pi inz}$$

I believe this can be proven from the partial fraction expansion of $cot(z)$, which I have found to be; $$cot(z)=\frac{1}{z}+\sum_{n \in \mathbb{Z}}^{'}\left( \frac{1}{z-n\pi}+ \frac{1}{n\pi}\right)$$ Where $'$ indicates that the $n=0$ term is excluded from the sum.

If I can prove the $k=1$ case the result clearly follows from differentating this case $k$ times. I've tried differentiating my expression for $cot(z)$ and have spent a lot of time fiddling around, but can't seem to get anywhere. A hint or a push in the right direction would be much appreciated.


Differentiating the expression for $cot(z)$, we have

$$-cosec^2(z)=\frac{-1}{z^2}+\sum_{n \in \mathbb{Z}}^{'}\frac{-1}{(z-n\pi)^2}=-\sum_{n \in \mathbb{Z}}\frac{1}{(z-n\pi)^2}$$

So, $$\pi^2cosec^2(\pi z)=\sum_{n \in \mathbb{Z}}\frac{1}{(z-n)^2}=\sum_{n \in \mathbb{Z}}\frac{1}{(z+n)^2}$$

Which is the left hand side of the desired expression for $k=1$. I still can't find a way to connect this to the right hand side, but the exponential is reminding me of fourier series.


I sketch a proof from Zagier's notes on modular forms.

Euler showed that, $$\frac{\pi}{\tan \pi z} = \lim_{N\rightarrow \infty}\sum_{n=-N}^N \frac{1}{z+n}$$ (Here we need the upper minus lower bound to be bounded so I set them to be the same magnitude bounding them to zero.)

Write the tangent out in terms of complex exponentials, $$ \frac{\pi}{\tan \pi z} = -\pi i \frac{e^{2\pi i z}+1}{1-e^{2\pi i z}}$$ The denominator allows a Taylor expansion provided $\Im(z) > 0$ (which I notice is different to the question where you restrict $\Re(z)>0$),

$$ \frac{\pi}{\tan \pi z} = -\pi i (e^{2\pi i z}+1)\sum_{n=0}^\infty e^{2\pi i n z}$$ Expand the bracket and tidy-up to learn, $$ \sum_{n=-\infty}^\infty \frac{1}{z+n}= -2\pi i\left[\frac{1}{2} + \sum_{n=1}^\infty e^{2\pi i n z}\right]$$ This is the key result you were looking for. I think it is already clear to you that next you should differentiate $k$ times and re-arrange.

  • $\begingroup$ Be careful with the order of summation in $\sum_n \frac{1}{z+n}$ and it is worth noting $f(z)=\frac{\pi^2}{\sin^2 \pi z} -\sum_n \frac{1}{(z+n)^2}$ is a $1$-periodic bounded entire function thus it is constant $ = f(i\infty)=0$ $\endgroup$ – reuns Jul 1 at 19:38

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