Product of $\prod_{n=0}^{k-1} \frac{-1}{(4n+4)(4n+3)}$

Question

Is there any way of expressing $$\prod_{n=0}^{k-1} \frac{-1}{(4n+4)(4n+3)}$$ as some special function? I can get factor some things out and get $$\frac{(-1)^k}{4^kk!}\prod_{n=0}^{k-1} \frac{1}{4n+3}$$ but this last thing is recalcitrant. It's something like a quadruple factorial, but I haven't seen much about those. This arises from the series solution to $y''+x^2y=0$.

Answer 1

If you have a product of the terms of an arithmetic progression, then you can factor out the "step size" and then use the gamma function. $$ \prod_{n=0}^{k-1}\left(an+b\right) = a^k\;\prod_{n=0}^{k-1}\left(n+\frac{b}{a}\right) $$ The gamma function satisfies the recurrence relation $$ \Gamma(z+1) = z\Gamma(z) $$ which in turn means $$ \Gamma(z+m+1) = z(z+1)(z+2)\cdots(z+m)\Gamma(z) = \Gamma(z)\prod_{n=0}^{m} (z+n) $$ Now set $m=k-1$ and $z=\frac{b}{a}$ und you get $$ \prod_{n=0}^{k-1}\left(an+b\right) = a^k\;\prod_{n=0}^{k-1}\left(n+\frac{b}{a}\right) =a^k\;\frac{\Gamma\left(k+\frac{b}{a}\right)}{\Gamma\left(\frac{b}{a}\right)} $$ In your specific case, this means $$ \prod_{n=0}^{k-1}\frac{-1}{(4n+4)(4n+3)} =\frac{(-1)^k\;}{4^k\;k!} \;\cdot\; \frac{\Gamma\left(\frac{3}{4}\right)}{4^k\;\Gamma\left(k+\frac{3}{4}\right)} =\frac{(-1)^k\;\Gamma\left(\frac{3}{4}\right)}{4^{2k}\;k!\;\Gamma\left(k+\frac{3}{4}\right)} $$