Summation of fractions of Gamma functions

Question

Recently I gave Mathematica the following input on the left hand side

$$ \sum_{n=0}^{\infty}\frac{\Gamma(a+n)}{\Gamma(b+n)}=\frac{\Gamma(a)\Gamma(b-a-1)}{\Gamma(b-1)\Gamma(b-a)}. $$

Can anyone explain to me what identities are needed to get the expression on the right hand side above?

Answer 1

Definition and properties of the Beta function: $$ \frac{\Gamma(a+n)}{\Gamma(b+n)}=\frac{1}{\Gamma(b-a)}B(a+n,b-a) = \frac{1}{\Gamma(b-a)}\int_{0}^{1}(1-x)^{b-a-1}x^{a+n-1}\,dx. $$ If you sum both sides on $n\geq 0$, you end up with:

$$ \sum_{n\geq 0}\frac{\Gamma(a+n)}{\Gamma(b+n)}=\frac{1}{\Gamma(b-a)}\int_{0}^{1}(1-x)^{b-a-2}x^{a-1}\,dx=\frac{B(a,b-a-1)}{\Gamma(b-a)}=\frac{\Gamma(a)\Gamma(b-a-1)}{\Gamma(b-1)\Gamma(b-a)}. $$

Answer 2

With Hypergeometric function \begin{align} \frac{\Gamma(a)\Gamma(b-a-1)}{\Gamma(b-1)\Gamma(b-a)} &=\frac{\Gamma(a)}{\Gamma(b)}{}_2F_1(1,a,b;1)\\ &=\frac{\Gamma(a)}{\Gamma(b)}\sum_{n=0}^{\infty}\frac{(1)_n(a)_n}{(b)_nn!}\\ &=\sum_{n=0}^{\infty}\frac{(1)_n}{n!}\frac{\Gamma(a)(a)_n}{\Gamma(b)(b)_n}\\ &=\sum_{n=0}^{\infty}\frac{\Gamma(a+n)}{\Gamma(b+n)} \end{align}