Does anyone have any ideas on howto verify $$\sum\_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$? 


I posted this question also on [math.stackexchange](https://math.stackexchange.com/questions/2364283/double-series-equals-gamma-function).


This is not an exercise from a book, but arises due to my research in the study of a probability density.



Double Series involving Gamma function
This problem can be reduced at least formally to a compact double integral, which might be easier to solve.


Starting with the integral representation for the Gamma function, we write the double sum as an integral of the square of the confluent hypergeometric function ${}\_1F\_1$, then apply analogue of Euler's transformation formula:
\begin{align}
&\sum\_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}}\\
&=3\int\_0^\infty e^{-3t}t^{3x-1}\sum\_{n,m=0}^\infty \frac{t^{n+m}}{\Gamma(n+1+x)\Gamma(m+1+x)}dt\\
&=\frac3{\Gamma^2(x+1)}\int\_0^\infty e^{-3t}t^{3x-1}~\left[{}\_1F\_1(1,x+1;t)\right]^2dt\\
&=\frac3{\Gamma^2(x+1)}\int\_0^\infty e^{-t}t^{3x-1}~\left[{}\_1F\_1(x,x+1;-t)\right]^2dt\\
&=\frac3{\Gamma^2(x+1)}\int\_0^\infty e^{-t}t^{3x-1}~\sum\_{n,m=0}^\infty\frac{(x)\_n(x)\_m}{(x+1)\_n(x+1)\_m}\frac{(-t)^{n+m}}{n!m!}dt\\
&=\frac{3x^2}{\Gamma^2(x+1)}\int\_0^\infty e^{-t}t^{3x-1}~\sum\_{n,m=0}^\infty\frac{1}{(x+n)(x+m)}\frac{(-t)^{n+m}}{n!m!}dt\\
&=\frac{3}{\Gamma^2(x)}\int\_0^\infty e^{-t}t^{3x-1}\int\_0^1 du\int\_0^1 dv\sum\_{n,m=0}^\infty\frac{(-t)^{n+m}u^{n+x-1}v^{m+x-1}}{n!m!} dt\\
&=\frac{3}{\Gamma^2(x)}\int\_0^\infty e^{-t}t^{3x-1}\int\_0^1 du\int\_0^1 dv~ e^{-t(u+v)}(uv)^{x-1}dt\\
&=\frac{3\Gamma(3x)}{\Gamma^2(x)}\int\_0^1\int\_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv.
\end{align}


This means that the initial problem is equivalent to
$$\int\_0^1\int\_0^1 \frac{u^{x-1} v^{x-1}}{(u+v+1)^{3 x}} dudv=\frac{\Gamma^3(x)}{3\Gamma(3x)}.
$$


