# 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

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Note that \begin{align} &\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty} {\pars{n + m}! \over n!\,m!\,n^{2}\,m^{2}}\pars{1 \over 2}^{n + m} = \sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}}\sum_{m = 1}^{\infty} {n + m \choose m}{1 \over m^{2}}\pars{1 \over 2}^{m} \\[5mm] = &\ \sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}}\sum_{m = 1}^{\infty} {-n - 1 \choose m}\pars{-1}^{m}{1 \over m^{2}}\pars{1 \over 2}^{m} \\[5mm] = &\ \sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}}\sum_{m = 1}^{\infty} {-n - 1 \choose m}\bracks{-\int_{0}^{1}\ln\pars{x}x^{m - 1}\,\dd x} \pars{-\,{1 \over 2}}^{m} \\[5mm] = &\ -\int_{0}^{1}{\ln\pars{x} \over x}\sum_{n = 1}^{\infty} {\pars{1/2}^{n} \over n^{2}}\bracks{\sum_{m = 1}^{\infty} {-n - 1 \choose m}\pars{-\,{x \over 2}}^{m}}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{\ln\pars{x} \over x}\sum_{n = 1}^{\infty} {\pars{1/2}^{n} \over n^{2}}\bracks{\pars{1 - {x \over 2}}^{-n - 1} - 1}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{\ln\pars{x} \over x}\braces{% \pars{1 - {x \over 2}}^{-1} \sum_{n = 1}^{\infty}{\bracks{1/\pars{2 - x}}^{\,n} \over n^{2}} - \sum_{n = 1}^{\infty}{\pars{1/2}^{n} \over n^{2}}}\,\dd x \\[5mm] = &\ -\int_{0}^{1}{\ln\pars{x} \over x}\bracks{% {2 \over 2 - x}\,\mrm{Li}_{2}\pars{1 \over 2 - x} - \,\mrm{Li}_{2}\pars{1 \over 2}}\,\dd x \approx 0.8494 \end{align}
To be continued$\ldots$
$$\mrm{Li}_{2}\pars{1 \over 2 - x} = {\pi^{2} \over 6} + \ln\pars{2 - x}\ln\pars{1 - x} - \mrm{Li}_{2}\pars{x - 1} - {1 \over 2}\,\ln^{2}\pars{2 - x}$$