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

[usumdelphini]

Homework Statement

sum this series:

S = \sum_{-\infty}^\infty \frac{1}{|x-kx_0|}

Homework Equations

The Attempt at a Solution

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

but I don't know how to evaluate these two sums :(

 

[D H]

Both of those series need to converge. Does either one do so?

 

[usumdelphini]

\sum_{k<A} \frac{1}{A-k} + \sum_{k>A} \frac{1}{k-A}

It looks like no, because they seems to behave as \sum \frac{1}{k} (?)

 

[D H]

Very good. You have to demonstrate this conclusively, but you are on the right track.

 

[usumdelphini]

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

 

[Infinitum]

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?

 

[usumdelphini]

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 ...

 

[Ray Vickson]

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
S(M,N) = \sum_{k=-M}^N \frac{1}{|x-k x_0|}.
The series converges if \lim_{M,N \rightarrow \infty} S(M,N) exists and is finite. Note, however, that the M and N limits are separate, not coupled. What you have shown is that \lim_{N \rightarrow \infty} S(N,N) exists and is finite, but that does not mean that S(M,N) converges.

RGV

 

[usumdelphini]

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 1/x + log(x-x_0) ?