Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)

$$\begin{align}\sum_{k=-\infty}^\infty\frac{1}{|x-kx_0|}&=\frac{1}{x}+\sum_{k=-\infty}^{-1}\frac{1}{|x-kx_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\\\\ & =\frac{1}{x}+\sum_{k'=1}^{\infty}\frac{1}{|x+k'x_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\end{align}$$
If $x\neq 0$, 
  $$=\frac{1}{x}+\sum_{x+k'x_0>0}\frac{1}{x+k'x_0}+\sum_{x+k'x_0<0}\frac{-1}{x+k'x_0}+\sum_{x-kx_0>0}\frac{1}{x-kx_0}+\sum_{x-kx_0<0}\frac{1}{kx_0-x}$$
  $$=\frac{1}{x}+\sum_{k>-\frac{x}{x_0}}\frac{1}{x+kx_0}-\sum_{k<-\frac{x}{x_0}}\frac{1}{x+kx_0}+\sum_{k<\frac{x}{x_0}}\frac{1}{x-kx_0}+\sum_{k>\frac{x}{x_0}}\frac{1}{kx_0-x}$$
  $$=\frac{1}{x}+\frac{1}{x_0}\left[\sum_{k>-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}-\sum_{k<-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}+\sum_{k<\frac{x}{x_0}}\frac{1}{-k+\frac{x}{x_0}}+\sum_{k>\frac{x}{x_0}}\frac{1}{k-\frac{x}{x_0}}\right]$$

and my prof's version (I'm not sure he could be so quick on the absolute value)
(original image)

$$\sum_{n=-\infty}^{+\infty}\frac{1}{|x-nx_0|}=\frac{1}{x}+\sum_{n=-\infty}^{-1}\frac{1}{|x-nx_0|}+\sum_{n=1}^\infty\frac{1}{|x-nx_0|}$$
  $$=\frac{1}{x}+\sum_{m=+\infty}^{+1}\underbrace{\frac{1}{x-nx_0}}_{\substack{\text{change variable }m=-n,\\\\ \Large \frac{1}{x+mx_0}}}+\sum_{n=1}^\infty\frac{1}{nx_0-x}$$
  $$=\sum_{n=1}^{+\infty}\underbrace{\frac{1}{nx_0-x}-\frac{1}{x+nx_0}}_{\Large\frac{x+nx_0-nx_0+x}{n^2x_0^2-x^2}}$$
  $$\frac{1}{x}+2x\sum_{n=1}^{+\infty}\frac{1}{n^2x_0^2-x^2}$$

Thanks!

As noted by some users, the series below is the one for the case
$$\sum_{k=-\infty}^{+\infty}\frac{1}{z-k}$$
i.e, there are no absolute values.
I scanned too fast but the last thing you have is this
$$\pi \cot(\pi z)=\frac 1 z+2z \sum_{n=1}^\infty \frac{1}{z^2-n^2} $$
One option is to use
$$\frac{{\sin \pi z}}{{\pi z}} = \prod\limits_{k = 1}^\infty  {\left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)} $$
Take logarithms and differentiate:
$$\eqalign{
  & \log \sin \pi z - \log \pi z = \log \prod\limits_{k = 1}^\infty  {\left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)}   \cr 
  & \log \sin \pi z - \log \pi z = \sum\limits_{k = 1}^\infty  {\log \left( {1 - \frac{{{z^2}}}{{{k^2}}}} \right)}   \cr 
  & \pi \frac{{\cos \pi z}}{{\sin \pi z}} - \frac{1}{z} = \sum\limits_{k = 1}^\infty  {\frac{{ - \frac{{2z}}{{{k^2}}}}}{{1 - \frac{{{z^2}}}{{{k^2}}}}}}   \cr 
  & \pi \cot \pi z = \frac{1}{z} - 2z\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^2} - {z^2}}}}  \cr} $$
$$\pi \cot \pi z = \frac{1}{z} + 2z\sum\limits_{k = 1}^\infty  {\frac{1}{{{z^2} - {k^2}}}} $$
I know virtually nothing about complex analysis, but what you start with is the decomposition of the cotangent into partial fractions in complex analysis, pretty much like one decomposes a polynomial with its roots, one does it with the singularities here.
