Sum of an infinite series involving product of three variables I would like to find a closed form expression for the sum:
$$\sum_{n_1=0}^\infty\sum_{n_2=0}^\infty\sum_{n=0}^\infty \binom{n_1+n}{n}\binom{n_2+n}{n}x_1^{n_1}x_2^{n_2}x^{n},$$
where the absolute values of $x_1,x_2,x$ as well as of their sums are assumed to be less than 1.
Any help is appreciated.
 A: By stars and bars we have
$$ \sum_{m\geq 0}\binom{n+m}{n}x^m = \frac{1}{(1-x)^{n+1}} $$
hence your triple series is simply given by
$$ \sum_{n\geq 0}\frac{x^n}{(1-x_1)^{n+1}(1-x_2)^{n+1}} = \color{red}{\frac{1}{(1-x_1)(1-x_2)-x}} $$
as soon as $x,x_1,x_2$ are close enough to the origin.
A: Use the fact that
$$\sum_{k\ge 0}\binom{k+n}nx^k=\frac1{(1-x)^{n+1}}$$
twice: as a formal power series
$$\begin{align*}
&\sum_{n_1\ge 0}\sum_{n_2\ge 0}\sum_{n\ge 0}\binom{n_1+n}n\binom{n_2+n}nx_1^{n_1}x_2^{n_2}x^n\\
&\qquad=\sum_{n\ge 0}x^n\sum_{n_1\ge 0}\binom{n_1+n}nx_1^{n_1}\sum_{n_2\ge 0}\binom{n_2+n}nx_2^{n_2}\\
&\qquad=\sum_{n\ge 0}\frac{x^n}{(1-x_2)^{n+1}}\sum_{n_1\ge 0}\binom{n_1+n}nx_1^{n_1}\\
&\qquad=\sum_{n\ge 0}\frac{x^n}{(1-x_1)^{n+1}(1-x_2)^{n+1}}\\
&\qquad=\frac1{(1-x_1)(1-x_2)}\sum_{n\ge 0}\left(\frac{x}{(1-x_1)(1-x_2)}\right)^n\\
&\qquad=\frac1{(1-x_1)(1-x_2)}\cdot\frac1{1-\frac{x}{(1-x_1)(1-x_2)}}\\
&\qquad=\frac1{(1-x_1)(1-x_2)-x}\;,
\end{align*}$$
and I’ll leave it to you to sort out the exact requirements for convergence.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \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}\,}\,}
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\begin{align}
&\sum_{n_{1}\ =\ 0}^{\infty}\,\sum_{n_{2}\ =\ 0}^{\infty}
\,\sum_{n\ =\ 0}^{\infty}
{n_{1} + n \choose n}{n_{2} + n \choose n}x_{1}^{n_{1}}x_{2}^{n_{2}}x^{n}
\\[5mm] = &\
\sum_{n\ =\ 0}^{\infty}x^{n}\sum_{n_{1}\ =\ 0}^{\infty}
{n_{1} + n \choose n_{1}}x_{1}^{n_{1}}\,\sum_{n_{2}\ =\ 0}^{\infty}
{n_{2} + n \choose n_{2}}x_{2}^{n_{2}}
\qquad\pars{~Binomial\ Symmetry~}
\\[5mm] = &\
\sum_{n\ =\ 0}^{\infty}x^{n}\sum_{n_{1}\ =\ 0}^{\infty}
\bracks{{-n - 1\choose n_{1}}\pars{-1}^{n_{1}}x_{1}^{n_{1}}}
\sum_{n_{2}\ =\ 0}^{\infty}
\bracks{{-n - 1 \choose n_{2}}\pars{-1}^{n_{2}}x_{2}^{n_{2}}}
\quad\pars{~Binomial\ Negating~}
\\[5mm] & =
\sum_{n = 0}^{\infty}x^{n}\
\bracks{1 + \pars{-x_{1}}}^{-n - 1}\
\bracks{1 + \pars{-x_{2}}}^{-n - 1}
\qquad\pars{~Newton\ Binomial\ Formula~}
\\[5mm] = &\
{1 \over \pars{1 - x_{1}}\pars{1 - x_{2}}}
\sum_{n = 0}^{\infty}\bracks{x \over \pars{1 - x_{1}}\pars{1 - x_{2}}}^{n}
\\[5mm] & =
{1 \over \pars{1 - x_{1}}\pars{1 - x_{2}}}
{1 \over 1 - x/\bracks{\pars{1 - x_{1}}\pars{1 - x_{2}}}}
\qquad\pars{~Newton\ Binomial\ Formula~}
\\[5mm] = &\
{1 \over \pars{1 - x_{1}}\pars{1 - x_{2}} - x} =
\bbx{\ds{1 \over x_{1}x_{2} - x_{1} - x_{2} - x + 1}}
\end{align}
