# Limit of product with prime numbers

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.

• What do you want to know about this product? – Rob Arthan Jan 14 '15 at 23:18
• @RobArthan The answer maybe... but we want to try to do this product before some of you give us the answer else there is no challenge... :D – ParaH2 Jan 14 '15 at 23:20
• Well... knowing the answer doesn't negate the challenge of proving that it is correct. It's pretty easy to guess what it must be, just by calculating it for the first couple of dozen primes in Excel... – Joffan Jan 14 '15 at 23:23
• I'm wondering if you can use $$\sum_{n=1}^\infty \frac{1}{n^s} = \prod_{p \in \Bbb{P}}\frac{p^s}{p^s-1}$$ somehow – graydad Jan 14 '15 at 23:23
• After numerical calculations with PARI/GP, I conjecture $\frac{7}{6}$. – Peter Jan 14 '15 at 23:25

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.

• How do you compute the ratio of the powers of zeta functions? – user2566092 Jan 15 '15 at 0:16
• @user2566092 the values of Zeta functions at even numbers are known explicitly; $\zeta(2n)$ is always a rational multiple of $\pi^{2n}$. – Steven Stadnicki Jan 15 '15 at 0:27
• Ah, there it is. Nicely done! – graydad Jan 15 '15 at 0:29

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 $n > 1$.