Prove two series are equal 
Prove $I=J$, where:
$$I=\left\{\frac{4}{\pi^6}\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=1}^{\infty}\frac{1}{n^2m^2\sqrt{n^2+m^2}}\left(\pi\frac{e^{\pi\sqrt{n^2+m^2}}+e^{-\pi\sqrt{n^2+m^2}}}{e^{\pi\sqrt{n^2+m^2}}-e^{-\pi\sqrt{n^2+m^2}}}-\frac{1}{\sqrt{n^2+m^2}}\right)\right\}^{-1},$$
and
$$J=12\pi^2\left(\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\right)^{-1}.$$

My atempt:  We have
\begin{align*}
\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+1)^2}+\frac{1}{(3n+2)^2}\right\}&=\displaystyle\sum_{n=1}^{\infty}\left\{\frac{1}{(3n-2)^2}+\frac{1}{(3n-1)^2}+\frac{1}{(3n)^2}\right\}-\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n)^2}\\
&=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}-\displaystyle\sum_{n=1}^{\infty}\frac{1}{(3n)^2}\\
&=\frac{8}{9}\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{4\pi^2}{27}\\
\end{align*}
and:
\begin{align*}
\displaystyle\sum_{n=0}^{\infty}\frac{1}{(3n+2)^2}&=4\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+4)^2}\\
&=4\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(6n+1)^2}+\frac{1}{(6n+4)^2}\right\}-4\displaystyle\sum_{n=1}^{\infty}\frac{1}{(6n+1)^2}\\
&=4\displaystyle\sum_{n=0}^{\infty}\left\{\frac{1}{(3n+1)^2}-\frac{1}{(6n+1)^2}\right\}\\
\end{align*}
So :
\begin{align*}
\frac{4\pi^2}{27}&=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(3n+1)^2}+4\displaystyle\sum_{n=0}^{\infty}\left(\frac{1}{(3n+1)^2}-\frac{1}{(6n+1)^2}\right)\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\\
&\implies \frac{1}{12\pi^2}\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}=\frac{1}{81}\\
&\implies 12\pi^2\left(\displaystyle\sum_{n=0}^{\infty}\frac{5}{(3n+1)^2}-\frac{4}{(6n+1)^2}\right)^{-1}=81\\
\end{align*}
Finally $ J=81$, but how to prove $I=81$?
Any help for how to calculate  $I$? Thanks in advance
 A: The first step is to recognize the fraction involving $e^{\pi\sqrt{n^2+m^2}}$ as $\coth (\pi\sqrt{n^2+m^2})$. Then we can use the Mittag-Leffler expansion of $\coth$,
$$
\pi\coth(\pi z) = \frac{1}{z} + 2 \sum_{n=1}^\infty \frac{z}{z^2+n^2}\ .
$$
Applying this to $z = \sqrt{m^2+n^2}$, the stuff inside the parantheses in $I$ can be rewritten as
$$
\frac{1}{\sqrt{m^2+n^2}} + 2\sum_{p=1}^\infty \frac{\sqrt{m^2+n^2}}{m^2+n^2+p^2} - \frac{1}{\sqrt{m^2+n^2}}\ .
$$
After canceling out the $1/\sqrt{m^2+n^2}$, note that the $\sqrt{m^2+n^2}$ factor outside of the parentheses also gets canceled.
Thus
$$
\frac{\pi^6}{4}I^{-1} = \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{2}{m^2n^2(m^2+n^2+p^2)}\ .
$$
To evaluate this sum, note that if we permute the dummy variables $m\mapsto n$, $n\mapsto p$, $p\mapsto m$, the summand can be rewritten as
$$
\frac{2}{n^2p^2(m^2+n^2+p^2)}\ .
$$
Permutting again turns the summand into
$$
\frac{2}{p^2m^2(m^2+n^2+p^2)}\ .
$$
Since relabeling the dummy variables clearly doesn't change the value of the series, we could add all three representations together and get
$$
\begin{align}
3\left(\frac{\pi^6}{8I}\right) &= \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \left(\frac{1}{m^2n^2(m^2+n^2+p^2)} +\right.\\
& \quad\left. \frac{1}{n^2p^2(m^2+n^2+p^2)} + \frac{1}{p^2m^2(m^2+n^2+p^2)}\right) \\
&= \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{p^2+m^2+n^2}{m^2n^2p^2(m^2+n^2+p^2)}\\
& = \sum_{m=1}^\infty \sum_{n=1}^\infty \sum_{p=1}^\infty \frac{1}{m^2n^2p^2} \\
& = \left(\sum_{m=1}^\infty \frac{1}{m^2}\right)^3 \\
& = \left(\frac{\pi^2}{6}\right)^3
\end{align}
$$
From this we can easily solve for $I = 81$.
