sequences and series - Infinite Sum of Falling Factorial and Power

According to Mathematica,

$$\sum_{k=0}^\infty \frac{(G+k)_{G-1}}{2^k}=2(G-1)!(2^{G}-1)$$

where

$$(G+k)_{G-1}=\frac{(G+k)!}{(G+k-G+1)!}=\frac{(G+k)!}{(k+1)!}$$

is the falling factorial. I would like to compute this analytically, but I have nothing I've been doing works. A proof by induction led me to a more complex summation, and I can split it or simplify the falling factorial. Is there any possible way to evaluate this without resorting to Mathematica? Any help and/or references would be greatly appreciated.

Answer

We have $$S=\sum_{k\geq0}\frac{\left(G+k\right)!}{\left(k+1\right)!}\left(\frac{1}{2}\right)^{k}=\left(G-1\right)!\sum_{k\geq0}\dbinom{G+k}{G-1}\left(\frac{1}{2}\right)^{k}
$$ and since holds $$\dbinom{G+k}{G-1}=\sum_{m=0}^{G-1}\dbinom{k+m}{m}
$$ we have, exchanging the sum with the series $$S=\left(G-1\right)!\sum_{m=0}^{G-1}\sum_{k\geq0}\dbinom{k+m}{m}\left(\frac{1}{2}\right)^{k}
$$ now note that $$\frac{\left(k+m\right)!}{m!}=\left(k+m\right)\left(k+m-1\right)\cdots\left(k+m-\left(k-1\right)\right)=\left(-1\right)^{k}\left(-\left(m+1\right)\right)_{k}$$ where $\left(x\right)_{k}$ is the Pochhammer' symbol, so by the generalized binomial theorem we have $$\sum_{k\geq0}\dbinom{k+m}{m}\left(\frac{1}{2}\right)^{k}=\sum_{k\geq0}\dbinom{-\left(m+1\right)}{k}\left(-\frac{1}{2}\right)^{k}$$ $$=\frac{1}{\left(1-\frac{1}{2}\right)^{m+1}}=2^{m+1}

$$ and finally $$\sum_{m=0}^{G-1}2^{m+1}=2\left(2^{G}-1\right)$$ so

$$\sum_{k\geq0}\frac{\left(G+k\right)!}{\left(k+1\right)!}\left(\frac{1}{2}\right)^{k}=2\left(G-1\right)!\left(2^{G}-1\right).$$