Infinite Sum of an Infinite Sum of an Infinite Sum

$A=\displaystyle \sum_{x=2}^{\infty} \left(\displaystyle \sum_{n=1}^{\infty} \left(\displaystyle \sum_{k=1}^{\infty} \dfrac{1}{x^nx^k}\right)\right)$

$A=\displaystyle \sum_{x=2}^{\infty} \left( \left(\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{x^n}\right)\left(\displaystyle \sum_{k=1}^{\infty}\dfrac{1}{x^k}\right)\right)$

$A=\displaystyle \sum_{x=2}^{\infty} \left(\displaystyle \sum_{n=1}^{\infty}\dfrac{1}{x^n}\right)^2$

$A=\displaystyle \sum_{x=2}^{\infty} \left(\cfrac{1}{1-\frac{1}{x}}-1\right)^2$

$A=\displaystyle \sum_{x=2}^{\infty} \left(\cfrac{1}{x-1}\right)^2$

$A=\displaystyle \sum_{x=1}^{\infty}\cfrac{1}{x^2}$

$A=\zeta{(2)}=\cfrac{\pi^2}{6}$

$\therefore m+n=8$

I'll see if I can provide an image later but this was my approach... Let x= 2 so that we have to calculate $\ A_2 = \displaystyle\sum_{n=1}^{\infty} \displaystyle\sum_{k=1}^{\infty} \frac{1}{2^{n+k}}$ Each term can be calculated in a table with say, the 'n' values across the top and the 'k' values along the side. So the table would read: $\ 1/4 \ ~~ \ 1/8 \ ~~ \ 1/16 \ ~~ \ 1/32 \ ~~ \ 1/64 \ ~ ...$ $\ 1/8 \ ~~ \ 1/16 \ ~~ \ 1/32 \ ~~ \ 1/64 \ ~~ \ 1/128 \ ~ ...$ $\ 1/16 \ ~~ \ 1/32 \ ~~ \ 1/64 \ ~~ \ 1/128 \ ~~ \ 1/256 \ ~ ...$ and so. If we let $\ S = 1+ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} +...$ then reading down the page we have: $\ A_2 = \frac{1}{4} S + \frac{1}{8} S + \frac{1}{16} S +... = \frac{1}{4}S (1 +\frac{1}{2}+ \frac{1}{4} + \frac{1}{8} +...) = \frac{1}{4}S^2 = \frac{1}{4} \times\ 2^2 = 1$ Now if we let x = n then we have $\ S = 1 + \frac{1}{n}+ \frac{1}{n^2 }+ \frac{1}{n^3} +... = \frac{1}{1-\frac{1}{n}}$ and the factor of 1/4 is generally $\frac{1}{n^2}$ so we have: $\ \displaystyle\sum_{x=2}^{\infty} \displaystyle\sum_{n=1}^{\infty} \displaystyle\sum_{k=1}^{\infty} \frac{1}{x^{n+k}} = \displaystyle\sum_{x=1}^{\infty} A_x = \displaystyle\sum_{x=2}^{\infty} \frac{1}{(x-1)^2} = \displaystyle\sum_{x=1}^{\infty} \frac{1}{x^2} = \frac{\pi^2}{6}$