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?