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I've these two series, and I would like a closed form:

$$ \sum_{k=-\infty}^{\infty} \frac{x+kx_0-h}{|x+kx_0|^3}$$

$$ 3\sum_{k=-\infty}^{\infty} \frac{(x+kx_0-h)(x+kx_0)^2}{|x+kx_0|^5} $$

Mathematica gives me these closed forms ($A=x_0$, $B=h$):

Sum[(x + k*A - B)/((abs (x + k*A))^3), {k, -Infinity, Infinity}]
  =   (1/(2 A^3 abs^3))(2 A PolyGamma[1, x/A] + 2 A PolyGamma[1, 1 - x/A] + 
     B PolyGamma[2, x/A] - B PolyGamma[2, 1 - x/A])
Sum[((x + k*A - B)*(x + k*A)^2)/(abs (x + k*A)^5), {k, -Infinity, Infinity }]
=  (1/(2 A^3 abs))(2 A PolyGamma[1, x/A] + 2 A PolyGamma[1, 1 - x/A] + B  
   PolyGamma[2, x/A] - B PolyGamma[2, 1 - x/A])

Now, first of all, when it writes $abs^3$ this should mean $|x|^3$ or $|x-kx_0|^3$? And at last, does this make sense? Is there an analytical way to reach this (or a better) conclusion?

Thanks!

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You may want to post your question here as well: mathematica.stackexchange.com –  math-visitor Jun 22 '12 at 5:46
    
Good idea! I didn't know about that! –  usumdelphini Jun 22 '12 at 5:48
1  
Are A and B the same as $x_0$ and $h$? I suspect you mistakenly used () instead of [] and Mathematica thinks abs is the name of a variable. –  Robert Israel Jun 22 '12 at 6:15
    
Added input code and yes $A=x_0$, $B=h$! –  usumdelphini Jun 22 '12 at 6:26
1  
Yes, that's your mistake. It should be Abs[...], not abs (...). –  Robert Israel Jun 22 '12 at 6:40
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1 Answer

If $0 < x < x_0$, $x + k x_0 > 0$ for $k \ge 0$ and $< 0$ for $k \le -1$, so your first sum is $$ - \sum_{k=-\infty}^{-1} \frac{x + k x_0 - h}{(x + k x_0)^3} + \sum_{k=0}^{\infty} \frac{x + k x_0 - h}{(x + k x_0)^3}$$ which Maple evaluates as $$ \frac{ h\Psi \left( 2,{\dfrac {x}{{ x_0}}} \right) +2 { x_0}\ \Psi \left( 1,{\dfrac {x}{ x_0}}\right) { x_0}}{2\, { x_0}^3}+ \frac{ h\Psi \left( 2,1-\dfrac{x}{ x_0} \right) -2\,{x_0}\,\Psi \left( 1,1-\dfrac{x}{x_0} \right) } {2\, {x_0}^3} $$

The second sum is just $3$ times the first if $x$ and $x_0$ are real, because $(x + k x_0)^2 = \left| x + k x_0 \right|^2$.

EDIT: If $x_0 > 0$ and $s$ is the fractional part of $x/x_0$ (so $0 < s < 1$ and $x/x_0 - s$ is an integer) then $x + k x_0 > 0$ for $k \ge n = s - x/x_0$ and $<0$ for $k \le n-1$, your first sum is $$ - \sum_{k=-\infty}^{n-1} \frac{x + k x_0 - h}{(x + k x_0)^3} + \sum_{k=n}^{\infty} \frac{x + k x_0 - h}{(x + k x_0)^3}$$ which Maple evaluates as $$ \dfrac{2 \Psi(1,s)}{ x_0^2} + \dfrac{h \Psi(2,s)}{x_0^3} + \dfrac{\pi^3 h \cot(\pi s) \csc(\pi s)^2}{x_0^3} - \dfrac{\pi^2 \csc(\pi s)^2}{x_0^2} $$

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Yes, but what about the closed form if $x$ can be every value, so not necessarily $0<x<x_0$? –  usumdelphini Jun 22 '12 at 16:23
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