m is a parameter of the expression. Whether it is integer or real has no impact on the solution.

The sum looks familiar from decades ago. I've no recollection though of where it arises.

I took m to be an integer because it is stated in your problem definition. :-)

Hi Michael,

Try FullSimplify on the result

Assuming[Element[m, Integers], 
 Sum[1/(n^2*(n - m)^2), {n, -Infinity, -1}] + Sum[1/(n^2*(n - m)^2), {n, 1, Infinity}]]

(*
(12 EulerGamma + m \[Pi]^2 + 12 PolyGamma[0, 1 - m] + 
  6 m PolyGamma[1, 1 - m])/(
 6 m^3) + (-12 EulerGamma + m \[Pi]^2 - 12 PolyGamma[0, 1 + m] + 
  6 m PolyGamma[1, 1 + m])/(6 m^3)
*)

FullSimplify@%

(*
(-9 + m^2 \[Pi]^2 + 3 m \[Pi] Csc[m \[Pi]]^2 (m \[Pi] + Sin[2 m \[Pi]]))/(3 m^4)
*)

I do, but couldnt find it there... At least not in the series section.

I actually did the first one (splitting it into 2 summations), but i cant see how seperating it to m<0, m=0, m>0 is equivalent to 'm is an integer'. i have tried a new approach based on your answer:

code:

Assuming[m \[Element] Integers, \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = \(-\[Infinity]\)\), \(-1\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\) + \!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(\[Infinity]\)]
\*FractionBox[\(1\), \(
\*SuperscriptBox[\(n\), \(2\)] 
\*SuperscriptBox[\((n - m)\), \(2\)]\)]\)]

and arrived to the following solotioun:

This might be the intended solution, but i am not certain. We have never used PolyGamma functions before, and although we dont really need to study it to get this result, i suspect it is not what the exercise was aiming for. I guess i am expecting a nice convergance to a compact form. So am i just paranoid or wolfram really misinterpet my quary?