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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?)

Dane Bouchie
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  • If $\phi(n)$ is the Euler totient function, then for primes $p$, we have $\frac{\phi(p)}{p} = \frac{p-1}{p} \to 1$ as $p \to \infty$. So, $\lim_{n \to \infty}\frac{\phi(n)}{n} \neq 0$. – JimmyK4542 Jan 01 '15 at 06:43
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    Yes, I was pointing out that your statement "$\lim_{n \to \infty}\frac{\phi(n)}{n} = 0$" wasn't correct. – JimmyK4542 Jan 01 '15 at 06:45
  • Ah, thanks for that, I read "lim inf ${\phi(n) \over n} = 0$" off wikipedia, but that must mean something else. – Dane Bouchie Jan 01 '15 at 06:47
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    If you haven't encountered $\liminf$ before, it is possibly unwise to start digging into this area yet. Review some real analysis first. It's not a lot, but a little will help. – Thomas Andrews Jan 01 '15 at 06:49
  • Already on it... – Dane Bouchie Jan 01 '15 at 06:51

3 Answers3

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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.

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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.)

Thomas Andrews
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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: $\displaystyle \prod_{i = 0}^{\infty}\dfrac{1}{1-\frac{1}{p_i^s}} = \sum_{n = 1}^{\infty}\dfrac{1}{n^s}$.

JimmyK4542
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