Infinite Product of Product of Sequence of n plus alpha over Sequence of n plus beta

Theorem

 * $\displaystyle \prod_{n \mathop \ge 1} \dfrac {\left({n + \alpha_1}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \cdots \left({n + \beta_k}\right)} = \dfrac {\Gamma \left({1 + \beta_1}\right) \cdots\Gamma \left({1 + \beta_1}\right)} {\Gamma \left({1 + \alpha_1}\right) \cdots\Gamma \left({1 + \alpha_k}\right)}$

where:
 * $\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$
 * none of the $\beta$s is a negative integer.

Proof
First we note that if any of the $\beta$s is a negative integer, the would have $0$ as its denominator, and so would be undefined.

We have from the Euler form of the Gamma function that:
 * $\Gamma \left({1 + \beta_i}\right) = \displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\left({1 + \beta_i}\right) \left({2 + \beta_i}\right) \cdots \left({m + 1 + \beta_i}\right)}$

and so the can be written as: