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Does anybody have an idea on how to evaluate the following double series? $$\sum_{n=1}^\infty \sum_{m=1}^\infty\frac{(-1)^{n+m}}{nm(n^2+m^2)}$$ Clearly this is a convergent series but I think it might be hard to find a closed expression for this.

nabla
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  • Is there any relation to this question asked yesterday, or is the similarity just a coincidence? – joriki Mar 14 '20 at 16:05
  • By "evaluate", do you mean finding a closed form or evaluating it numerically? – joriki Mar 14 '20 at 16:09
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    @joriki Yes, there is! I saw the similar question yesterday and then I somehow started thinking about this more symmetrical problem and how one could find a closed form for this expression. I don't think that this helps to answer the original question, that is why I opened a new post. – nabla Mar 14 '20 at 16:24
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    You may be interested in this question. – Paul Enta Mar 14 '20 at 16:30
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    An approach similar to my answer in the linked question suggests a first step: \begin{eqnarray} \sum_{m=1}^\infty \sum_{n=1}^\infty\frac{(-1)^{n+m}}{nm(n^2+m^2)} &=&\sum_{m=1}^\infty\left[\frac{(-1)^m}{m}\sum_{n=1}^\infty\frac{(-1)^n}{n(n^2+m^2)}\right]\ &=&2\sum_{m=1}^{\infty}\frac{\operatorname{Re}\psi(mi)-\operatorname{Re}\psi(\tfrac m2i)}{m^2}\ &=&2\sum_{m\geq1}\frac{\operatorname{Re}\psi(mi)}{m^2} -\frac12\sum_{m\geq1}\frac{\operatorname{Re}\psi((m-\tfrac12)i)}{(m-\tfrac12)^2}. \end{eqnarray} A closed form for this seems unlikely, however. – Servaes Mar 19 '20 at 12:56

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