Sum of a Simple Series: Solving for S using Homework Equations

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Homework Statement



sum this series:

S = [itex]\sum_{-\infty}^\infty \frac{1}{|x-kx_0|}[/itex]

Homework Equations


The Attempt at a Solution



[itex]S = \sum_{|x-kx_0|<0} \frac{1}{kx_0-x}+\sum_{|x-kx_0|>0}\frac{1}{x-kx_0}[/itex]

but I don't know how to evaluate these two sums :(
 
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[itex]\sum_{k<A} \frac{1}{A-k} + \sum_{k>A} \frac{1}{k-A}[/itex]

It looks like no, because they seems to behave as [itex]\sum \frac{1}{k}[/itex] (?)
 
Voilà, I think this is going to work, so does the series converge? view attachement
 

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usumdelphini said:
Voilà, I think this is going to work, so does the series converge? view attachement

Based on that work, what would happen if n2x02 = x2??

Could the series converge, then?
 
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According to this simulation with matlab, x=1, x_0=2, the series converges, but differently than that written in the last calculation (previous attachement) In blue the 1/x + 2x sum ...
 

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usumdelphini said:
According to this simulation with matlab, x=1, x_0=2, the series converges, but differently than that written in the last calculation (previous attachement) In blue the 1/x + 2x sum ...

I don't think the series converges in the usual sense. Llet
[tex]S(M,N) = \sum_{k=-M}^N \frac{1}{|x-k x_0|}.[/tex]
The series converges if [itex]\lim_{M,N \rightarrow \infty} S(M,N)[/itex] exists and is finite. Note, however, that the M and N limits are separate, not coupled. What you have shown is that [itex]\lim_{N \rightarrow \infty} S(N,N)[/itex] exists and is finite, but that does not mean that S(M,N) converges.

RGV
 
Yes, actually I'm looking for an expression explicit in x of the series, so, whether if the series has limit or not, it can still be written like - for instance - something like [itex]1/x + log(x-x_0)[/itex] ?