# Evaluating Sum involving binomial coefficients and powers

I would like to evaluate the double sum $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} \dfrac{(n+m)!}{n!m!n^2 m^2}\left(\dfrac{1}{2}\right)^{n+m}$. My starting point was to consider $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} \dfrac{(n+m)!}{n!m!}x^n y^m = \dfrac{1}{1 -x -y} - \dfrac{1}{1-x} - \dfrac{1}{1-y} + 1$ $\;\;\forall\;\; |x|+|y|<1\;\;$ All what is left is to divide by $xy$ then integrate with respect to $x$ and then with respect to $y$ (process should be repeated twice) finally set $x = y = \frac{1}{2}$. I am however stuck in evaluating the resulting integrals. I expect logarithmic and polylogarithmic functions to show up in the final result. I would appreciate if you can help me formulating the value of this sum. Thanks for your help...

• The proposed method: $$\sum_{n=1}^{\infty}\,\sum_{m=1}^{\infty}\,\frac{(n+m)!}{n!\,\,m!}\,\frac{x^n y^m}{n\,m} = \int_{0}^{x}\int_{0}^{y}\,\frac{1}{u\,v}\left(\frac{1}{1-u-v}-\frac{1}{1-u}-\frac{1}{1-v}+1\right)\,dv\,du$$ Could work to evaluate: $$\sum_{n=1}^{\infty}\,\sum_{m=1}^{\infty}\,\frac{(n+m)!}{n!\,\,m!\,\,n\,\, m}\left(\frac{1}{2}\right)^{n+m} = \zeta(2)+\log^2(2)$$ And, it seems super-hard to perform another double integral step using same method again! – Hazem Orabi Feb 15 '17 at 11:30
• wolframalpha gives 0.849413 as the answer. Try using Cauchy product? – hypergeometric Feb 15 '17 at 15:03
• I want a closed form expression, I don't want numerical values...Thanks Hazem Orabi for your efforts... – Hmath Feb 15 '17 at 17:49

To be continued$\ldots$
$$\mrm{Li}_{2}\pars{1 \over 2 - x} = {\pi^{2} \over 6} + \ln\pars{2 - x}\ln\pars{1 - x} - \mrm{Li}_{2}\pars{x - 1} - {1 \over 2}\,\ln^{2}\pars{2 - x}$$