Double Series equals Gamma function I want to show $$\sum_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)}\cdot \frac{1}{3^{n+m+3x-1}} = \Gamma(x)$$ for $x>0$ and I don't know how.
 A: All credit goes to @Nemo. As he pointed out in mathoverflow.net, \begin{align*}
&\sum\limits_{n,m=0}^\infty \frac{\Gamma(n+m+3x)}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}} \\
&= \int\limits_0^\infty t^{n+m+3x-1}e^{-t} \sum\limits_{n,m=0}^\infty \frac{1}{\Gamma(n+1+x)\Gamma(m+1+x)} \frac{1}{3^{n+m+3x-1}}\\
&\overset{t=3u}{=} 3\int\limits_{0}^{\infty}u^{3x-1} e^{-3u} \left(\sum\limits_{n=0}^\infty \frac{u^n}{\Gamma(n+1+x)}\right)^2 du .
\end{align*} 
As written in the Wikipedia article to the Incomplete Gamma Function, $$\sum\limits_{n=0}^\infty \frac{u^n}{\Gamma(n+1+x)} = \frac{u^{-x}}{\Gamma(x)} e^u \int\limits_0^u v^{x-1} e^{-v} dv,$$ so we get 
\begin{align*}
&\frac{1}{(\Gamma(x))^2} \cdot 3\int\limits_{0}^{\infty} u^{x-1}e^{-u} \left(\int\limits_0^u v^{x-1} e^{-v} dv\right)^2 du\\
&= \frac{1}{(\Gamma(x))^2} \cdot 3 \int\limits_0^{\Gamma(x)} s^2ds\\
&= \frac{1}{(\Gamma(x))^2} \cdot (\Gamma(x))^3\\
&= \Gamma(x),
\end{align*} where we used the substitution $s(u):= \int\limits_0^u w^{x-1}e^{-w} dw$.
