Series involving gamma functions

[polygamma]

$\displaystyle \sum_{n={\bf 0}}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{n! \Gamma(c+n)} \ \ c-a-b > 0$

 

[polygamma]

I don't like this problem. It's really just Gauss' hypergeometric theorem in disguise.$ \displaystyle F(a,b;c;z) = \sum_{n=0}^{\infty} \frac{(a)_{n} (b)_{n}}{(c)_{n}} \frac{z_{n}}{n!} = \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n) \Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c+n)} \frac{z^{n}}{n!} $

$\displaystyle = \frac{1}{B(b,c-b)} \int_{0}^{1} x^{b-1} (1-x)^{c-b-1} (1-zx) ^{-a} \ dx \ \ |z|<1 \ \text{or} \ z=1$ let $z=1$$\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n) \Gamma(c)}{\Gamma(a) \Gamma(b) \Gamma(c+n)} \frac{1}{n!}
= \frac{B(b,c-a-b)}{B(b,c-b)} = \frac{\Gamma(b) \Gamma(c-a-b) \Gamma(c)}{\Gamma(b) \Gamma(c-b) \Gamma(c-a)} $so $\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n)}{n!\Gamma(c+n)} = \frac{\Gamma(a) \Gamma(b) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}$