1
$\begingroup$

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!

  • 0
    Now, the error lies in the last equal sign since the numerator of each fraction in the last sum should be $2k$ and not $1$.2012-06-18

3 Answers 3

1

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.

  • 0
    @usumdelphini: Glad to help.2012-06-18
1

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.

  • 0
    I think this is the intended sum. Cheers!2012-06-18
1

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}

  • 0
    Perfect, I arrive there. Then, how can I avoid the separation of the sum due to the study of the absolute value?2012-06-17