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As a follow up of this nice question I am interested in

$$ S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2} $$

Furthermore, I would be also very grateful for a solution to

$$ S_2=\sum_{m=1}^{\infty}\sum_{n=m+1}^{\infty}\frac{ 1}{m n\left(m^2-n^2\right)^2} $$

Following my answer in the question mentioned above and the numerical experiments of @Vladimir Reshetnikov it's very likely that at least

$$ S_1+S_2 = \frac{a}{b}\pi^6 $$

I think both sums may be evaluated by using partial fraction decomposition and the integral representation of the Polygamma function but I don't know how exactly and I guess there could be a much more efficient route.

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    $\begingroup$ The inner summation in the first line is summed from $n=0$ to $n-1$. This needs to be adjusted to make sense. As it should also be adjusted in the problem from whence this is first presented. $\endgroup$
    – Leucippus
    May 22, 2015 at 14:06
  • $\begingroup$ Sorry for the cross-edits, just making sure the title was covered too $\endgroup$
    – abiessu
    May 22, 2015 at 14:13
  • $\begingroup$ no problem. now everthing should be fine... :) $\endgroup$
    – tired
    May 22, 2015 at 14:13
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    $\begingroup$ Where does $m$ start? $\endgroup$
    – Alex M.
    May 22, 2015 at 14:15
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    $\begingroup$ @RenatoFaraone Makes no difference (for $S_1$), since the inner sum is an empty sum for $m = 1$. $\endgroup$ May 22, 2015 at 14:28

2 Answers 2

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Clearly, $S_1$=$S_2$ (this can be shown by reversing the order of summation, as was noted above). Using $$ \frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{ (m+n)^2-(m-n)^2}{4 m^2 n^2\left(m^2-n^2\right)^2} $$ we get $$ S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}, $$ and after reversing the order of summation in the first sum $$ S_1=\frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m-n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\ \frac{1}{4}\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{4}\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2}. \qquad\qquad (1) $$

Let's introduce a third sum $$ S_3=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m^2 n^2\left(m+n\right)^2} =\sum_{n=1}^{\infty}\sum_{m=n+1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}=\\ \frac{1}{2}\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{ 1}{m^2 n^2\left(m+n\right)^2}-\frac{1}{8}\zeta(6). $$ Using An Infinite Double Summation $\sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$? we get $$ S_3=\frac{1}{2}\cdot\frac{1}{3}\zeta(6)-\frac{1}{8}\zeta(6)=\frac{1}{24}\zeta(6).\qquad\qquad\qquad (2) $$ From (1) and (2) we get

$$ S_1=\frac{1}{4}\cdot\frac{1}{3}\zeta(6)-\frac{1}{4}\cdot\frac{1}{24}\zeta(6)=\frac{7}{96}\zeta(6)=\frac{7}{96}\frac{\pi^6}{945}=\frac{\pi^6}{12960} $$

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  • $\begingroup$ excellent work (+1) , i was failing after your formula one back then....! $\endgroup$
    – tired
    Oct 18, 2015 at 10:48
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Numerically, I get $$ S_1+S_2 = 0.14836252987273216621 $$ which agrees with $$ \frac{\pi^6}{6480} $$ Also numerically, $$ S_1 = 0.074181264936366083104 \\ S_2 = 0.074181264936366083104 $$ are seemingly equal.

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  • $\begingroup$ very well! multiplying this by $8\pi$ and adding $\pi^7/70$gives us the result suggested by @Vladimir Reshetnikov $\endgroup$
    – tired
    May 22, 2015 at 15:31
  • $\begingroup$ Do you have an idea where this symmerty come from? I'm not getting my head around it :/ $\endgroup$
    – tired
    May 22, 2015 at 15:54
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    $\begingroup$ Reverse the order of summation in $S_2$ and see what you get. $\endgroup$
    – GEdgar
    May 22, 2015 at 16:41
  • $\begingroup$ thanks, oh man today is not my brightest day! $\endgroup$
    – tired
    May 22, 2015 at 17:25

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