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How to calculate$$\sum_{n=-\infty}^{\infty}{\sum_{m=-\infty}^{\infty}{\frac{\left(-1\right)^n}{\left(6m\right)^2+\left(6n+1\right)^2}}}.$$Follow this,we first get $$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {6k + 1} \right)}^2} + {l^2}}}} = \frac{{{l^2}}}{{{l^4} + 1}} - \frac{1}{{{l^2} + 1}} - \frac{\pi }{{12l}}\left( {\frac{{\sinh \frac{{\pi l}}{6}}}{{\cosh \frac{{\pi l}}{6} + \frac{{\sqrt 3 }}{2}}} + \frac{{\sinh \frac{{\pi l}}{6}}}{{ - \cosh \frac{{\pi l}}{6} + \frac{{\sqrt 3 }}{2}}}} \right).$$But it seems difficult to continue!

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    $\begingroup$ Why are people voting to close? $\endgroup$ Feb 2, 2017 at 22:22
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    $\begingroup$ I downvoted because I think that the OP should explain the context. How did you come to that summation? Why are there those 6's and not a general parameter $a$? What do you mean by "calculate"? $\endgroup$
    – user40023
    Feb 3, 2017 at 15:37
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    $\begingroup$ @Fry, OP would have to be a mindreader to understand that that's what you found wrong with the question. Can I encourage you (and others) to help posters by leaving a comment when you downvote? (or even instead of making that downvote?) $\endgroup$ Feb 3, 2017 at 22:57

1 Answer 1

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The paper Two-dimensional series evaluations via the elliptic functions of Ramanujan and Jacobi deals exactly with double sums of this kind and shows how to evaluate them in terms of elliptic functions.

The book Lattice sums then and now gives alternative methods to calculate lattice sums.

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  • $\begingroup$ Thank you, Lattice sums are series about $(-1)^{m+n}$, my question is about $(-1)^n$,I hope Lattice sums' method can work well. $\endgroup$
    – user165013
    Feb 7, 2017 at 9:42

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