What is the product of $p_i-1 \over p_i$

Question

I am trying to find the value of $\prod_{i=0}^{\infty}{p_i-1 \over p_i}$ = ${\lim_{x \to \infty}} {\phi(p_x!) \over p_x!}$ Where $p_x!$ is the $x$th primorial, and $p_i$ is the $i$th prime number.

I guess I can honestly say I have no idea where to start, other than just iteriating it manually (around $0.25$ maybe?)

Answer 1

Hint:-

Note that, $$\displaystyle\prod_{i=1}^\infty\left(1-\dfrac{1}{p_i}\right)=\dfrac{1}{\zeta(1)}$$

Where $\zeta$ is the Riemann-Zeta Function.

Answer 2

If $a_i$ is a positive sequence, and $\prod (1-a_i)$ converges to a non-zero value, then there is a general theorem which says that $\sum a_i$ must converge.

But $\sum_{p<x} \frac{1}{p} = O(\log \log x)$ is know to diverge.

In the above case, this means that the product converges to zero (which is often thought of as the equivalent of saying the product diverges.)

Answer 3

Consider the reciprocal $\displaystyle\prod_{i = 0}^{\infty}\dfrac{p_i}{p_i-1} = \prod_{i = 0}^{\infty}\dfrac{1}{1-\frac{1}{p_i}} = \prod_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\dfrac{1}{p_i^j}$.

This can be written out as:

$\left(\dfrac{1}{1}+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\cdots\right)\left(\dfrac{1}{1}+\dfrac{1}{3^1}+\dfrac{1}{3^2}+\cdots\right)\left(\dfrac{1}{1}+\dfrac{1}{5^1}+\dfrac{1}{5^2}+\cdots\right) \cdots$

Each integer $n$ has a unique prime factorization $n = p_1^{e_1}p_2^{e_2} \cdots p_r^{e_r}$. So, when you multiply out that product, there is exactly one $\dfrac{1}{n} = \dfrac{1}{p_1^{e_1}}\dfrac{1}{p_2^{e_2}} \cdots \dfrac{1}{p_r^{e_r}}$ term for each integer $n$.

Therefore, $\displaystyle\prod_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\dfrac{1}{p_i^j} = \sum_{n = 1}^{\infty}\dfrac{1}{n}$ which diverges to $+\infty$. Hence, the reciprocal $\displaystyle\prod_{i = 0}^{\infty}\dfrac{p_i-1}{p_i}$ is $0$.

Note: This is a specific case of the more general Euler Product: $\displaystyle \prod_{i = 0}^{\infty}\dfrac{1}{1-\frac{1}{p_i^s}} = \sum_{n = 1}^{\infty}\dfrac{1}{n^s}$.