# Rational function sequence product

I want to know a closed formula for $$\prod_{m = 0}^{n} \frac{m^2 + a}{m^2 + a + 1},$$ a being any given complex.

When the exponent is 1, it's pretty trivial, because of cancellations, but with other positive integers, I can't figure this out.

-
Do you have any reason to think there is a closed form? –  Gerry Myerson May 8 '11 at 5:18
There is a closed form in terms of gamma functions, but I'm not too sure if it's terribly useful. –  Ｊ. Ｍ. May 8 '11 at 5:19

(might as well...)

We can factor the numerator and denominator like so:

$$\prod_{m = 0}^{n} \frac{m^2 + a}{m^2 + a + 1}=\frac{\left(\prod\limits_{m = 0}^{n}\left(m+i\sqrt{a}\right)\right)\left(\prod\limits_{m = 0}^{n}\left(m-i\sqrt{a}\right)\right)}{\left(\prod\limits_{m = 0}^{n}\left(m+i\sqrt{1+a}\right)\right)\left(\prod\limits_{m = 0}^{n}\left(m-i\sqrt{1+a}\right)\right)}$$

and recall the definition for Pochhammer symbols:

$$(a)_{n+1}=\prod_{m = 0}^{n}(a+m)$$

which yields the expression

$$\prod_{m = 0}^{n} \frac{m^2 + a}{m^2 + a + 1}=\frac{\left(i\sqrt{a}\right)_{n+1}\left(-i\sqrt{a}\right)_{n+1}}{\left(i\sqrt{1+a}\right)_{n+1}\left(-i\sqrt{1+a}\right)_{n+1}}$$

or, using the recursion relation $(a)_{m+n}=(a)_m (a+m)_n$,

$$\frac{\left(i\sqrt{a}\right)_{n+1}\left(-i\sqrt{a}\right)_{n+1}}{\left(i\sqrt{1+a}\right)_{n+1}\left(-i\sqrt{1+a}\right)_{n+1}}=\frac{a\left(1+i\sqrt{a}\right)_n\left(1-i\sqrt{a}\right)_n}{(1+a)\left(1+i\sqrt{1+a}\right)_n\left(1-i\sqrt{1+a}\right)_n}$$

We could then use the relation between gamma functions and Pochhammer symbols:

$$(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}$$

such that the whole mess is in terms of gamma functions like in Robert's answer. From that, we could then apply the reflection relation

$$\Gamma(1+iy)\Gamma(1-iy)=\pi y\;\mathrm{csch}(\pi y)$$

to have an expression with the minimum number of gamma functions and hyperbolic functions; I will skip that since Robert seems to have covered that, and will instead draw attention to the fact that if $a$ is of the form $a=-\ell^2$ or $a=-\ell^2-1$ for $\ell$ a positive integer, the gamma functions become singular and we have to stick with the Pochhammer symbol representation. In this respect, one could use the special formula for Pochhammer symbols with negative integer argument:

$$(-\ell)_n=(-1)^n n!\binom{\ell}{n}=(-1)^n (\ell-n+1)_n$$

so for instance, if $a=-\ell^2$,

$$\frac{\left(i\sqrt{-\ell^2}\right)_{n+1}\left(-i\sqrt{-\ell^2}\right)_{n+1}}{\left(i\sqrt{1-\ell^2}\right)_{n+1}\left(-i\sqrt{1-\ell^2}\right)_{n+1}}=\frac{\left(-\ell\right)_{n+1}\left(\ell\right)_{n+1}}{\left(-\sqrt{\ell^2-1}\right)_{n+1}\left(\sqrt{\ell^2-1}\right)_{n+1}}=\frac{(-1)^{n+1}\left(\ell-n\right)_{n+1}\left(\ell\right)_{n+1}}{\left(-\sqrt{\ell^2-1}\right)_{n+1}\left(\sqrt{\ell^2-1}\right)_{n+1}}$$

and similarly for the case $a=-\ell^2-1$...

-
$$\frac{\Gamma\left(\sqrt{-a-1}\right) \Gamma\left(-\sqrt{-a-1}\right)}{\Gamma\left(\sqrt{-a}\right) \Gamma\left(-\sqrt{-a}\right)} \frac{\Gamma\left(n+1+\sqrt{-a}\right) \Gamma\left(n+1-\sqrt{-a}\right)}{\Gamma\left(n+1+\sqrt{-a-1}\right) \Gamma\left(n+1-\sqrt{-a-1}\right)}$$