# series involving gamma functions

#### Random Variable

##### Well-known member
MHB Math Helper
$\displaystyle \sum_{n={\bf 0}}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{n! \Gamma(c+n)} \ \ c-a-b > 0$

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#### Random Variable

##### Well-known member
MHB Math Helper
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 $\displaystyle \sum_{n=0}^{\infty} \frac{\Gamma(a+n) \Gamma(b+n)}{n!\Gamma(c+n)} = \frac{\Gamma(a) \Gamma(b) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}$