Compute $\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$ Compute the product: $$\prod_{k=5}^{\infty}\frac{k^4-17k^2+16}{k^4-8k^2+16}$$
I was able to factor in the following manner:
$$ \frac{k^4-17k^2+16}{k^4-8k^2+16}=\frac{(k-1)(k+1)(k-4)(k+4)}{(k-2)^2(k+2)^2}$$
but what should I do now?
 A: Take the partial products up to $N$ and use the fact that $$\prod_{k=5}^N \frac{k-1}{k-2} = \frac{4}{3} \cdot \frac{5}{4} \cdot ... \cdot \frac{N-1}{N-2} = \frac{N-1}{3},$$ and similarly $$\prod_{k=5}^N \frac{k+1}{k+2} = \frac{6}{N+2}, \quad \prod_{k=5}^N \frac{k-4}{k-2} = \frac{2}{(N-2)(N-3)}, \quad \prod_{k=5}^N \frac{k+4}{k+2} = \frac{(N+4)(N+3)}{7 \cdot 8}.$$ So $$\prod_{k=5}^N \frac{k^4 - 17k^2 + 16}{k^4 - 8k^2 + 16} = \frac{(N-1)(N+3)(N+4)}{14 (N-3)(N-2)(N+2)}.$$
A: As you noted we have $$\prod_{k\geq5}\frac{k^{4}-17k^{2}+16}{k^{4}-8k+16}=\prod_{k\geq0}\frac{\left(k+4\right)\left(k+6\right)\left(k+1\right)\left(k+9\right)}{\left(k+3\right)^{2}\left(k+7\right)^{2}}$$ and now using the identity $$\prod_{k\geq0}\frac{\left(k+a\right)\left(k+b\right)\left(k+c\right)\left(k+d\right)}{\left(k+e\right)\left(k+f\right)\left(k+g\right)\left(k+h\right)}=\frac{\Gamma\left(e\right)\Gamma\left(f\right)\Gamma\left(g\right)\Gamma\left(h\right)}{\Gamma\left(a\right)\Gamma\left(b\right)\Gamma\left(c\right)\Gamma\left(d\right)},\,a+b+c+d=e+f+g+h$$ which follows from the definition of gamma as an infinite product, we have $$\prod_{k\geq5}\frac{k^{4}-17k^{2}+16}{k^{4}-8k+16}=\frac{\Gamma^{2}\left(3\right)\Gamma^{2}\left(7\right)}{\Gamma\left(4\right)\Gamma\left(6\right)\Gamma\left(1\right)\Gamma\left(9\right)}=\color{red}{\frac{1}{14}}$$ as wanted.
