Product of Primes

Question

Let $\mathbb{P}$ denote the set of prime numbers. How would one evaluate $$\prod_{p\in \mathbb{P}}\frac{p-1}{p}$$ I do not think that the fact that $$\prod_{n=2}^{\infty}\frac{n-1}{n}=\lim_{n\to\infty}\frac{1}{n}=0$$ can be applied, but it is worth noting. Additionally, if we take the log, we obtain $$\log\left(\prod_{p\in \mathbb{P}}\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log\left(\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log(p-1)-\sum_{p\in\mathbb{P}}\log(p)$$

Answer 1

For $Re(s)> 1$ we have the Euler product $$ \frac{1}{\zeta(s)}=\prod_p (1-p^{-s}). $$ Taking the limit $s\to 1$ shows that the infinite product $\prod_p (1-\frac{1}{p})$ is $0$, since $\lim_{s\to 1}\zeta(s)=\infty$.

A second possibility is to use that the product $\displaystyle\prod_{n=1}^{\infty} (1- a_n)$ is non-zero if and only if $\sum_{n=1}^{\infty} a_n < \infty$, provided $a_i\in (0,1)$. But since $\sum_p\frac{1}{p}=\infty$, taking $a_i=\frac{1}{p_i}$ gives that the infinite product is zero.

Answer 2

You could also use Mertens' 3rd theorem which states that $$ \lim_{n\to\infty}\log n\prod_{p\le n}\left(1-\frac1p\right)=e^{-\gamma}\;. $$