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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)

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


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The last thing you got is the expasion of the cotangent! – Pedro Tamaroff Jun 17 '12 at 21:19
Both versions are a little strange, since the sums $\sum_{k=1}^{\infty} |\frac{1}{x-kx_0}|$ diverge, just like the harmonic series. – Cocopuffs Jun 17 '12 at 21:22
There is a sign error near the middle of the second block of computations, which seems to be due to the misconception that $\sum\limits_{n=+\infty}^1a_n$ would be $-\sum\limits_{n=1}^{+\infty}a_n$, while it is in fact $+\sum\limits_{n=1}^{+\infty}a_n$. – Did Jun 17 '12 at 21:32
I've typed the images in LaTeX to improve readability, and I did my best to stay true to the original images, but I may have inadvertently introduced changes. Please double-check that I've captured your intended text correctly. – Zev Chonoles Jun 17 '12 at 21:35
Perfectly captured. And thank you about this. So, the sum must diverge as one could aspect due to the similarity with the harmonic series? Actually, I'm absolutely with you on the error in my prof's proof, that I consider wrong. – usumdelphini Jun 17 '12 at 22:26
up vote 1 down vote accepted

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.

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So, what is wrong in this proof of convergence? $$ S=\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_{0}|}=\frac{1}{x_{0}}\left(\sum_{k<x/‌​x_{0}}\frac{1}{x/x_{0}-k}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0}}\right)=\frac{1}{x_{‌​0}}\left(\sum_{-k'<x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x_{0}}\frac{1}{k-x/x_{0‌​}}\right)=\frac{1}{x_{0}}\left(\sum_{k'>-x/x_{0}}\frac{1}{x/x_{0}+k'}+\sum_{k>x/x‌​_{0}}\frac{1}{k-x/x_{0}}\right)\leq\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{‌​x/x_{0}+k}+\frac{1}{k-x/x_{0}}\right) =\frac{1}{x_{0}}\sum_{k>x/x_{0}}\left(\frac{1}{k^{2}-x^{2}/x_{0}^{2}}\right) $$ – usumdelphini Jun 18 '12 at 4:17
@usumdelphini: Notice that $1/(k+a)+1/(k-a) = 2k/(k^2-a^2)$. – user26872 Jun 18 '12 at 4:24
yep, sorry. So it diverges. Thanks! – usumdelphini Jun 18 '12 at 4:26
@usumdelphini: Glad to help. – user26872 Jun 18 '12 at 4:27

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}

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Perfect, I arrive there. Then, how can I avoid the separation of the sum due to the study of the absolute value? – usumdelphini Jun 17 '12 at 22:30

As noted by some users, the series below is the one for the case


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.

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I think this is the intended sum. Cheers! – user26872 Jun 18 '12 at 3:45

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