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

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## 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 left hand side 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 right hand side can be written as:

 $\displaystyle$  $\displaystyle \dfrac {\left({\displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_1} m^{1 + \beta_2} \cdots m^{1 + \beta_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)} }\right)} {\left({\displaystyle \lim_{m \mathop \to \infty} \dfrac {m^{1 + \alpha_1} m^{1 + \alpha_2} \cdots m^{1 + \alpha_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} }\right)}$ $\displaystyle$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \dfrac {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right) m^k m^{\beta_1 + \beta_2 + \cdots + \beta_k} \left({m!}\right)^k} {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right) m^k m^{\alpha_1 + \alpha_2 + \cdots + \alpha_k} \left({m!}\right)^k}$ $\displaystyle$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \dfrac {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right) m^{\beta_1 + \beta_2 + \cdots + \beta_k} } {\displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right) m^{\alpha_1 + \alpha_2 + \cdots + \alpha_k} }$ simplifying slightly $\displaystyle$ $=$ $\displaystyle \lim_{m \mathop \to \infty} \displaystyle \prod_{1 \mathop \le n \mathop \le m + 1} \dfrac {\left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)}$ as $\alpha_1 + \cdots + \alpha_k = \beta_1 + \cdots + \beta_k$ $\displaystyle$ $=$ $\displaystyle \displaystyle \prod_{n \mathop \ge 1} \dfrac {\left({n + \alpha_1}\right) \left({n + \alpha_2}\right) \cdots \left({n + \alpha_k}\right)} {\left({n + \beta_1}\right) \left({n + \beta_2}\right) \cdots \left({n + \beta_k}\right)}$

$\blacksquare$