TITLE: Sum of $\Gamma(n+a) / \Gamma(n+b)$
QUESTION [13 upvotes]: If $a$ and $b$ are positive real numbers, such that $b > a + 1$, can we find the sum $$\sum_{n=0}^{\infty} \frac{\Gamma(n+a)}{\Gamma(n+b)}?$$ 
For example I have found that $$\sum_{n=0}^{\infty} \frac{\Gamma(n+3/2)}{\Gamma(n+3)} = \sqrt{\pi} = \Gamma(1/2)$$ and $$\sum_{n=0}^{\infty} \frac{\Gamma(n+4/3)}{\Gamma(n+4)} = \frac{3}{10}\Gamma(4/3)$$ but no general rule.

REPLY [8 votes]: Recalling the $\beta$ function 

$$ \mathrm{\beta}(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt ,  $$

we have
$$ \frac{\Gamma(n+a)}{\Gamma(n+b)} =  \frac{\beta(n+a,b-a)}{\Gamma(b-a)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1} t^{n+a-1}(1-t)^{b-a-1}dt   $$
$$ \implies \sum_{n=0}^{\infty} \frac{\Gamma(n+a)}{\Gamma(n+b)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1} t^{a-1}(1-t)^{b-a-2}dt = \frac{\beta(a,b-a-1)}{\Gamma(b-a)} $$
$$= {\frac {\Gamma  \left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b-1 \right) }}
=  {\frac {(b-1)\Gamma\left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b \right) }}.$$