Limit of product with prime numbers

Solution 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.

Solution 2:

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 $n > 1$.