Summation question?

Question

I have following summation that I want to implement using Mathematica:

Sum[
 If[m == 0 && n == 0
  , Nothing
  , (1/z^2 + (1/(z - (m 2 π + n 2 π τ))^2 - 
         1/((m 2 π + 
              n 2 π τ))^2) /. {z -> π τ}) - (1/
        z^2 + (1/(z - (m 2 π + n 2 π τ))^2 - 
         1/((m 2 π + n 2 π τ))^2) /. z -> π) // 
   Simplify
  ], {m, -∞, ∞}, {n, -∞, ∞}
 ]

Above code returns the following result:

$-\frac{2 \left(\tau ^2-1\right)}{\pi ^2 \tau ^2}$

However, if I remove the If command in the above summation, Mathematica doesn't do the summation anymore, but I know the above expression is finite when m=0&&n=0. What's going on here?

Answer 1

For Timing reasons, its not a good idea to Simplify inside a sum, because Sum has Attribute HoldFirst, meaning that each summand is simplified with actual numbers n,m. But this sum has to be evaluated algebraically (or by table lookup).

Simplification reveals two problems, removal of 1/m^2, 1/n^2 at infinity and

(1/z^2 + (1/(z - (m 2 \[Pi] + n 2 \[Pi] \[Tau]))^2 - 
   1/((m 2 \[Pi] + 
        n 2 \[Pi] \[Tau]))^2) /. {z -> \[Pi] \[Tau]}) - (1/
  z^2 + (1/(z - (m 2 \[Pi] + n 2 \[Pi] \[Tau]))^2 - 
   1/((m 2 \[Pi] + n 2 \[Pi] \[Tau]))^2) /. 
z -> \[Pi]) // Simplify

$$\frac{1}{(2 \pi m+2 \pi n \tau -\pi \tau )^2}-\frac{1}{(-2 \pi m-2 \pi n \tau +\pi )^2}+\frac{\frac{1}{\tau ^2}-1}{\pi ^2}$$

a constant term, that, in a double sum, occurs inifintely often.

Probably, there is a wrong sign of the compensating constant for the limit $n,m\to 0$

  \[Pi]^2 FullSimplify@ExpandAll[(-1 + 1/\[Tau]^2)/\[Pi]^2 - 
    1/(\[Pi] - 2 m \[Pi] - 2 n \[Pi] \[Tau])^2 + 
  1/(2 m \[Pi] - \[Pi] \[Tau] + 2 n \[Pi] \[Tau])^2 /. {n ->  0}] 

$$\frac{1}{(\tau -2 m)^2}-\frac{1}{(1-2 m)^2}+\frac{1}{\tau ^2}-1$$