TAOCP 1.2.5 Exercise 17

Section 1.2.5: Permutations and Factorials

Exercise 17. [**] [HM20] Prove that the infinite product

$$ \prod_{n \ge 1}\frac{(n+\alpha_1)\cdots(n+\alpha_k)}{(n+\beta_1)\cdots(n+\beta_k)} $$

equals

$$ \frac{\Gamma(1+\beta_1)\cdots\Gamma(1+\beta_k)}{\Gamma(1+\alpha_1)\cdots\Gamma(1+\alpha_k)}, $$

if $\alpha_1+\cdots+\alpha_k=\beta_1+\cdots+\beta_k$ and if none of the $\beta$'s is a negative integer.

Verified: yes
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Let

$$ P_N=\prod_{n=1}^{N}\frac{(n+\alpha_1)\cdots(n+\alpha_k)} {(n+\beta_1)\cdots(n+\beta_k)}. $$

The infinite product is, by definition,

$$ \prod_{n\ge1}\frac{(n+\alpha_1)\cdots(n+\alpha_k)} {(n+\beta_1)\cdots(n+\beta_k)} =\lim_{N\to\infty}P_N, $$

provided the limit exists.

First assume that none of the numbers $1+\alpha_i$ and $1+\beta_i$ is a nonpositive integer, so that all relevant gamma values are finite.

Euler's limit formula gives

$$ \Gamma(1+\alpha) = \lim_{N\to\infty} \frac{N!,N^\alpha} {\prod_{n=1}^{N}(n+\alpha)}. $$

Hence

$$ \prod_{n=1}^{N}(n+\alpha) = \frac{N!,N^\alpha}{\Gamma(1+\alpha)} \bigl(1+o(1)\bigr), \qquad N\to\infty. $$

Similarly,

$$ \prod_{n=1}^{N}(n+\beta) = \frac{N!,N^\beta}{\Gamma(1+\beta)} \bigl(1+o(1)\bigr). $$

Substituting these formulas into $P_N$, we obtain

$$ P_N = \prod_{i=1}^{k} \frac{ \dfrac{N!,N^{\alpha_i}}{\Gamma(1+\alpha_i)} (1+o(1)) }{ \dfrac{N!,N^{\beta_i}}{\Gamma(1+\beta_i)} (1+o(1)) }. $$

Since there are only finitely many factors, the product of finitely many terms of the form $1+o(1)$ is again $1+o(1)$. Therefore

$$ P_N = \left( \prod_{i=1}^{k} \frac{\Gamma(1+\beta_i)} {\Gamma(1+\alpha_i)} \right) N^{\sum_{i=1}^{k}(\alpha_i-\beta_i)} (1+o(1)). $$

By hypothesis,

$$ \sum_{i=1}^{k}\alpha_i = \sum_{i=1}^{k}\beta_i, $$

so

$$ N^{\sum_{i=1}^{k}(\alpha_i-\beta_i)}=1. $$

Hence

$$ P_N = \left( \prod_{i=1}^{k} \frac{\Gamma(1+\beta_i)} {\Gamma(1+\alpha_i)} \right) (1+o(1)). $$

Taking $N\to\infty$, we find that $P_N$ converges and

$$ \lim_{N\to\infty}P_N = \frac{\Gamma(1+\beta_1)\cdots\Gamma(1+\beta_k)} {\Gamma(1+\alpha_1)\cdots\Gamma(1+\alpha_k)}. $$

This proves the formula when no $\alpha_i$ is a negative integer.

Now suppose some $\alpha_j=-m$ with $m\ge1$. Then the factor

$(n+\alpha_j)$ vanishes when $n=m$, so $P_N=0$ for all $N\ge m$. Thus the infinite product is $0$.

On the right-hand side, $1+\alpha_j=1-m$ is a nonpositive integer, so $\Gamma(1+\alpha_j)$ has a pole. Therefore

$$ \frac1{\Gamma(1+\alpha_j)}=0, $$

and consequently

$$ \frac{\Gamma(1+\beta_1)\cdots\Gamma(1+\beta_k)} {\Gamma(1+\alpha_1)\cdots\Gamma(1+\alpha_k)} =0. $$

Hence the stated identity also holds in this case.

The hypothesis that no $\beta_i$ is a negative integer ensures that none of the factors $\Gamma(1+\beta_i)$ has a pole and, equivalently, that no denominator factor $n+\beta_i$ ever vanishes.

Therefore

$$ \prod_{n \ge 1}\frac{(n+\alpha_1)\cdots(n+\alpha_k)} {(n+\beta_1)\cdots(n+\beta_k)} = \frac{\Gamma(1+\beta_1)\cdots\Gamma(1+\beta_k)} {\Gamma(1+\alpha_1)\cdots\Gamma(1+\alpha_k)}. $$

$\square$