Today I wanted to evaluate this sum using Mathematica:

$$f(a,b):=\sum_{(k,l)\in\mathbb{Z}^2,\ l\ne0}\frac{1}{(2 i k a \pi + l b) (2 i (k - l) a \pi + l b)}$$

Sum[ If[ l == 0, 0, 
1/((2 I k a \[Pi] + l b) (2 I (k - l) a \[Pi] + l b))
], {l, -Infinity, Infinity}, {k, -Infinity, Infinity}]    

And the result I am given is 0.

However, if I give specific values to $a, b$ and then evaluate the sum numerically, I obtain values different to $0$, as, for $a= 30, b=2$:

Sum[If[l == 0, 0, 
N[1/((2 I k 30 \[Pi] + l 2) (2 I (k - l) 30 \[Pi] + l 2))]]
, {l, -1000, 1000}, {k, -1000, 1000}]

Obtaining the output:

0.000137911 + 2.53623*10^-6 I

Plotting the sum with $b=2$:

If[l == 0, 0, 
a^2/((2 I k a \[Pi] + l 2) (2 I (k - l) a \[Pi] + l 2))]
, {l, -70, 70}, {k, -70, 70}]], {a, 15, 100}]

Seems that $|f|=O(\frac{1}{a^2}), f \ne 0$.

What it wrong with the symbolic calculation?

  • $\begingroup$ Indeed, easy to notice that at a=0 the sum transforms into Sum[If[l == 0, 0, 1/l^2], {l, -Infinity, Infinity}, {k, -Infinity, Infinity}]. Since Sum[If[l == 0, 0, 1/l^2], {l, -Infinity, Infinity}]==\[Pi]^2/3 the first sum is infinitely large, rather than zero. $\endgroup$ Commented Dec 4, 2018 at 13:30
  • $\begingroup$ @AlexeiBoulbitch you mean $\pi^2/6$ ? $\endgroup$ Commented Dec 4, 2018 at 13:53
  • $\begingroup$ @Alexei Boulbitch : The case $a=0$ is special. $\endgroup$
    – user64494
    Commented Dec 4, 2018 at 15:06
  • $\begingroup$ That is true, we may assume that both $a, b$ are positive reals. In spite of that, why is Mathematica giving 0 as an output and how can I get it to solve the sun correctly? $\endgroup$ Commented Dec 4, 2018 at 15:10
  • $\begingroup$ IMPO, this is rather a math question than a Mathematica question. The convergence of double series depending on parameters is a complex topic. Ask your question at math.stackexchange.com and/or mathoverflow.net . $\endgroup$
    – user64494
    Commented Dec 4, 2018 at 15:48

1 Answer 1


Let us put a=1/(2*\[Pi]) and b=1 in order to work with a concrete series. Next, let us use n instead of yourl (The use of l is not convenient because of its similarity to the imaginary unit I and the unit 1.). True series are absolutely convergent series (see https://en.wikipedia.org/wiki/Series_(mathematics) ,especially the Absolute convergence section, as a first reading). Let us calculate (notice that the terms of the series are complex-valued)

FullSimplify[ComplexExpand[Abs[1/((I*k + n)*(I*(k - n) + n))]], 
 Assumptions -> k \[Element] Reals && n \[Element] Reals]

1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)]

The sum of 1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)] over all the integer values of k and all the integer values of n but zero can be split into two sums. Let us consider one of these

Sum[1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)], {n, 1, Infinity}, {k, -Infinity, Infinity}]

.Mathematica returns it unevaluated on my comp. However, arguments similar to used in the integral test show that this sum is of the same quantity as the double improper integral

Integrate[1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)],{n,1,Infinity},{k,-Infinity,Infinity}]

See a typical plot

Plot3D[1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)], {k, 10, 11}, {n, 9,  12}]

enter image description here

to this end.

The above integral is divergent. This is clear if one switches to polar coordinates by (the r multiplier is the Jacobean)

1/Sqrt[(k^2 + n^2) (k^2 - 2 k n + 2 n^2)]*r /. {n -> r*Sin[\[Phi]], k -> r*Cos[\[Phi]]}//FullSimplify

$$\frac{\sqrt{2} r}{\sqrt{-r^4 (2 \sin (2 \phi )+\cos (2 \phi )-3)}} $$

Therefore, the sum of the series under consideration has no sense. Maybe, its sum exists as some generalized sum. In view of it the result

Sum[1/((I*k + n)*(I*(k - n) + n)), {n, 1, Infinity}, {k, -Infinity, Infinity}]


is not reliable, in particular, the sum may depend on the order of the terms. It should be noticed that

Sum[1/((I*k + n)*(I*(k - n) + n)),  {k, -Infinity,  Infinity},{n, 1, Infinity}]

returns unevaluated.


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