Sum of $\Gamma(n+a) / \Gamma(n+b)$ 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.
 A: Recalling the $\beta$ function 

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

we have
$$ \frac{\Gamma(n+a)}{\Gamma(n+b)} =  \frac{\beta(n+a,b-a)}{\Gamma(b-a)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1} t^{n+a-1}(1-t)^{b-a-1}dt   $$
$$ \implies \sum_{n=0}^{\infty} \frac{\Gamma(n+a)}{\Gamma(n+b)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1} t^{a-1}(1-t)^{b-a-2}dt = \frac{\beta(a,b-a-1)}{\Gamma(b-a)} $$
$$= {\frac {\Gamma  \left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b-1 \right) }}
=  {\frac {(b-1)\Gamma\left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b \right) }}.$$
A: in general
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}\Gamma{(b+k)}}{k!\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$$
note that
$$I=\int_{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-x)^{-a}dx=B(b,c-a-b)=\dfrac{\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}}$$
and we have
$$(1-x)^{-a}=\sum_{k=0}^{\infty}\binom{a-1+k}{k}x^k$$
then 
\begin{align*}I&=\sum_{k=0}^{\infty}\binom{a-1+k}{k}\int_{0}^{1}x^{b+k-1}(1-x)^{c-b-1}dx=\sum_{k=0}^{\infty}\dfrac{(a-1+k)!}{(a-1)!k!}B(b+k,c-b)\\
&=\sum_{k=0}^{\infty}
\dfrac{\Gamma{(a+k)}}{\Gamma{(a)}k!}\cdot\dfrac{\Gamma{(b+k)}\Gamma{(c-b)}}{\Gamma{(c+k)}}
\end{align*}
so
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}\Gamma{(b+k)}}{k!\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(b)}\Gamma{(c-a-b)}}{\Gamma{(c-a)}\Gamma{(c-b)}}$$
so let $b=1$
then
$$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}}{\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(c-a-1)}}{\Gamma{(c-1)}}$$
