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

From ProofWiki
Jump to navigation Jump to search


It has been suggested that this page be renamed.
In particular: If someone can come up with a better title than this (perhaps by finding that someone put their name to it) then please go to it.
To discuss this page in more detail, feel free to use the talk page.


Theorem

$\ds \prod_{n \mathop \ge 1} \dfrac {\paren {n + \alpha_1} \cdots \paren {n + \alpha_k} } {\paren {n + \beta_1} \cdots \paren {n + \beta_k} } = \dfrac {\map \Gamma {1 + \beta_1} \cdots \map \Gamma {1 + \beta_k} } {\map \Gamma {1 + \alpha_1} \cdots \map \Gamma {1 + \alpha_k} }$

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:

$\map \Gamma {1 + \beta_i} = \ds \lim_{m \mathop \to \infty} \dfrac {m^{1 + \beta_i} m!} {\paren {1 + \beta_i} \paren {2 + \beta_i} \cdots \paren {m + 1 + \beta_i} }$

and so the right hand side can be written as:

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

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Infinite_Product_of_Product_of_Sequence_of_n_plus_alpha_over_Sequence_of_n_plus_beta&oldid=528338"