# Sum of a simple series

1. Jun 16, 2012

### usumdelphini

1. The problem statement, all variables and given/known data

sum this series:

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

2. Relevant equations

3. 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 :(

2. Jun 16, 2012

### D H

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

3. Jun 16, 2012

### 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 $\sum \frac{1}{k}$ (?)

4. Jun 16, 2012

### D H

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

5. Jun 16, 2012

### usumdelphini

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

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6. Jun 17, 2012

### Infinitum

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

Could the series converge, then?

Last edited: Jun 17, 2012
7. Jun 17, 2012

### 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 ....

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8. Jun 17, 2012

### Ray Vickson

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

9. Jun 17, 2012

### 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 $1/x + log(x-x_0)$ ?