TITLE: Limit of product with prime numbers
QUESTION [16 upvotes]: A friend found a calculus problem in an old box with a lot of math exercises, but we don't have the answer to one of them. If you could help us with a hint it would be nice! The question is:  what is the limit of the following infinite product?
$$
\prod_{p \in \mathbb{P}} \frac{p^4+1}{p^4-1}
$$
Here $\mathbb{P}$ is the set of prime numbers.

REPLY [4 votes]: More generally:
Let
$R_n
=\prod_{p \in \mathbb{P}} \frac{p^n+1}{p^n-1}
$.
Since
$\zeta(s)
=\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p \in \Bbb{P}}\frac{p^s}{p^s-1}
$
and
$p^n+1
=\frac{p^{2n}-1}{p^n-1}
$,
$\begin{array}\\
R_n
&=\prod_{p \in \mathbb{P}} \frac{\frac{p^{2n}-1}{p^n-1}}{p^n-1}\\
&=\prod_{p \in \mathbb{P}} \frac{p^{2n}-1}{(p^n-1)^2}\\
&=\prod_{p \in \mathbb{P}} \frac{p^{2n}-1}{p^{2n}}\frac{p^{2n}}{(p^n-1)^2}\\
&=\prod_{p \in \mathbb{P}} \frac{p^{2n}-1}{p^{2n}}\prod_{p \in \mathbb{P}}\frac{p^{2n}}{(p^n-1)^2}\\
&=\frac1{\zeta(2n)}\left(\prod_{p \in \mathbb{P}}\frac{p^{n}}{p^n-1}\right)^2\\
&=\frac{\zeta^2(n)}{\zeta(2n)}\\
\end{array}
$
For even integer $n$,
this is a rational number,
but
$R_n =\frac{\zeta^2(n)}{\zeta(2n)}$
 is also true
for real $n > 1$.