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!

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}
