TITLE: Summation of $\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)}$
QUESTION [8 upvotes]: For $x>0$ and $y>x+1$, how do we prove that $$\sum\limits_{n=1}^{\infty} \frac{x(x+1) \cdots (x+n-1)}{y(y+1) \cdots (y+n-1)} = \frac{x}{y-x-1}$$

REPLY [9 votes]: This is based on Robert Smith's observation above and Robin Chapman's beta trick in some previous problem.
$\frac{\Gamma{(x+n)}\Gamma{(y-x)}}{\Gamma{(y+n)}} = \int_{0}^{1} t^{x+n-1} (1-t)^{y-x-1} dt$, 
summing over $n$  we get,
$\Gamma{(y-x)}\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \int_{0}^{1} \sum_{n \geq 1} t^{x+n-1} (1-t)^{y-x-1} dt = \int_{0}^{1} t^x (1-t)^{y-x-2}dt, $
or,
$\sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{1}{\Gamma(y-x)}\int_{0}^{1} t^{x+1-1}(1-t)^{y-x-1-1}dt = \frac{\Gamma{(x+1)}\Gamma{(y-x-1)}}{\Gamma{(y-x)}\Gamma(y)} = \frac{\Gamma{(x+1)}}{(y-x-1)\Gamma(y)}$
and hence,
$\frac{\Gamma(y)}{\Gamma(x)} \sum_{n \geq 1}\frac{\Gamma{(x+n)}}{\Gamma{(y+n)}} = \frac{\Gamma(x+1)}{(y-x-1)\Gamma(x)} = \frac{x}{y-x-1}.$