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.
On
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 $n > 1$.
Hint: Using Gray Dad's suggestion, along with the fact that $p^4+1=\dfrac{p^8-1}{p^4-1}$ , one should easily arrive at the conclusion that $P=\dfrac{\zeta^2(4)}{\zeta(8)}=\dfrac76$ , thus confirming Peter's numerical result.