# Summation of fractions of Gamma functions

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?

• What have you already tried? Try adding your thoughts to the question in order to get the most appropriate answers Commented Oct 30, 2018 at 16:44

## 2 Answers

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:

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

• Wow, that was pretty fast. Thanks a lot! Commented Oct 30, 2018 at 16:42

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}