Evaluation of $\prod_{k=1}^{\infty}\frac{a+k^2}{b+k^2}$ While playing around with the question The convergence of a sequence with infinite products, I found Mathematica to give me the result
$$
\prod\limits_{k=1}^{\infty}\frac{a+k^2}{b+k^2}
=
\frac{\sqrt{b}\,\sinh\left(\pi\sqrt{a}\right)}{\sqrt{a}\,\sinh\left(\pi\sqrt{b}\right)}
$$
Given the pretty answer, I thought someone might have a nice method to solve it.
Additionally, higher even powers instead of $2$ give similar results, so hopefully a solution to this would shed some light on how to evaluate those problems too.
The following question is probably relvant, but I haven't found a way to translate my problem into it: Infinite Product $\prod_{n=1}^\infty\left(1+\frac1{\pi^2n^2}\right)$

Well, 
$$\prod_{k=1}^{+\infty}\frac{a+k^2}{b+k^2}=\frac{\prod_{k=1}^{+\infty}\left(1+\frac{a}{k^2}\right)}{\prod_{k=1}^{+\infty}\left(1+\frac{b}{k^2}\right)}\tag{1}$$
hence you just need to prove that:
$$\prod_{k=1}^{+\infty}\left(1+\frac{c}{k^2}\right)=\frac{\sinh(\pi\sqrt{c})}{\pi\sqrt{c}}\tag{2}$$
that follows from the Weierstrass product for the hyperbolic sine function:
$$\frac{\sinh(z)}{z}=\prod_{k\geq 1}\left(1+\frac{z^2}{\pi^2 k^2}\right)\tag{3}$$
by simply taking $z=\pi\sqrt{c}$.
