Revival of Double Summation

Let $\mathcal{S}=\displaystyle\sum_{m=1}^{\infty}\displaystyle \sum_{n=1}^{\infty} \frac{1}{m[m(4m^2+4n^2-4m+1)-n(2m+2n-1)]}$

Simplifying we obtan

$\mathcal{S}=\displaystyle\sum_{m=1}^{\infty}\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{(2m^2-m)(2m^2-m+2n^2-n)}..........(1)$

$\therefore \mathcal{S}=\displaystyle\sum_{n=1}^{\infty}\displaystyle \sum_{m=1}^{\infty} \dfrac{1}{(2n^2-n)(2m^2-m+2n^2-n)}$ (Replacing $m$ by $n$ and $n$ by $m$ )

$\therefore \mathcal{S}=\displaystyle\sum_{m=1}^{\infty}\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{(2n^2-n)(2m^2-m+2n^2-n)}$ (Exchanging the order of summation) $..........(2)$

Adding $(1)$ and $(2)$

$2\mathcal{S}=\displaystyle\sum_{m=1}^{\infty}\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{(2m^2-m)(2n^2-n)}$

$\therefore \mathcal{S}=\dfrac{1}{2}\left( \displaystyle\sum_{m=1}^{\infty}\dfrac{1}{2m^2-m}\right)^{2}$

$\therefore \mathcal{S}=\dfrac{1}{2}\left( 2\left( 1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}...\right)\right)^{2}$

$\therefore \mathcal{S}=\boxed{2\ln^{2} 2}$