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...      
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
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$\ldots$

\begin{align}
\totald{}{x}\mrm{Li}_{2}\pars{1 \over 2 - x} & =
-\totald{}{x}\int_{0}^{1/\pars{2 - x}}{\ln\pars{1 - t} \over t}\,\dd t =
-\,{\ln\pars{1 - 1/\bracks{2 - x}} \over 1/\pars{2 - x}}
\,{1 \over \pars{2 - x}^{2}}
\\[5mm] & =
-\,{\ln\pars{1 - x} \over 2 - x} + {\ln\pars{2 - x} \over 2 - x}
\end{align}

\begin{align}
\mrm{Li}_{2}\pars{1 \over 2 - x} - \,\mrm{Li}_{2}\pars{1} & =
-\int_{1}^{x}{\ln\pars{1 - t} \over 2 - t}\,\dd t +
\int_{1}^{x}{\ln\pars{2 - t} \over 2 - t}\,\dd t
\\[5mm] & =
\int_{0}^{1 - x}{\ln\pars{t} \over 1 + t}\,\dd t -
\left.{1 \over 2}\,\ln^{2}\pars{2 - t}\,\right\vert_{\ t\ =\ 1}^{\ t\ =\ x}
\\[5mm] & =
-\int_{0}^{x - 1}{\ln\pars{-t} \over 1 - t}\,\dd t -
{1 \over 2}\,\ln^{2}\pars{2 - x}
\\[5mm] & =
\ln\pars{2 - x}\ln\pars{1 - x} - \int_{0}^{x - 1}\mrm{Li}_{2}'\pars{t}\,\dd t -
{1 \over 2}\,\ln^{2}\pars{2 - x}
\end{align}

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