Sum of $\Gamma(n+a) / \Gamma(n+b)$

Question

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.

Answer 1

Recalling the $\beta$ function

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

we have

$$ \frac{\Gamma(n+a)}{\Gamma(n+b)} = \frac{\beta(n+a,b-a)}{\Gamma(b-a)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1} t^{n+a-1}(1-t)^{b-a-1}dt $$

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

$$= {\frac {\Gamma \left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b-1 \right) }} = {\frac {(b-1)\Gamma\left( a \right) }{ \left( b-a-1 \right) \Gamma \left( b \right) }}.$$

Answer 2

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 $$\sum_{k=0}^{\infty}\dfrac{\Gamma{(a+k)}}{\Gamma{(c+k)}}=\dfrac{\Gamma{(a)}\Gamma{(c-a-1)}}{\Gamma{(c-1)}}$$