Q:

Lipschitz Summation Formula

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.
EDIT:
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.

A:

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.

