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!
 A: 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.
A: Maybe, you could try something with this
\begin{equation}
\sum_{k=-\infty}^{\infty}\frac{1}{|x-kx_0|} = \sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=0}^{\infty}\frac{1}{|x-kx_0|} = 
\end{equation}
\begin{equation}
= \frac{1}{|x|}+\sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=1}^{\infty}\frac{1}{|x-kx_0|} = \frac{1}{|x|}+\sum_{k=1}^{\infty}\bigg(\frac{1}{|x+kx_0|} +\frac{1}{|x-kx_0|}\bigg)
\end{equation}
A: The sum is divergent. 
Let's assume $x/x_0\notin\mathbb{Z}$ to avoid a trivial divergence. 
Note that for $k\ne 0$
$$\begin{eqnarray*}
|x-k x_0| &=& |k| |x/k-x_0| \\
&\le& |k|(|x/k| + |x_0|) \\
&\le& |k|(|x|+|x_0|).
\end{eqnarray*}$$
We have used the triangle inequality and the fact that 
$1/|k| \le 1$ for $|k|\ge 1$. 
Thus,
$$\begin{eqnarray*}
\sum_{k=-\infty}^\infty \frac{1}{|x-k x_0|}
&\ge& \frac{1}{|x|} + \frac{2}{|x| + |x_0|} \sum_{k=1}^\infty \frac{1}{k}.
\end{eqnarray*}$$
The sum diverges since the harmonic series diverges.
I suspect the sum you are actually interested in is the one dealt with by @PeterTamaroff.
Addendum: 
I have tracked down the sign error. 
Assuming $x/x_0 < 1$, so that $|n x_0 -x| = n x_0 - x$, 
the relevant term in the sum is
$$\frac{1}{n x_0 -x} + \frac{1}{n x_0 + x} 
= \frac{2 x_0 n}{n^2 x_0^2 - x^2} \sim \frac{1}{n}.$$
Note the relative sign is plus, not minus. 
