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 $$\log\left(\prod_{p\in \mathbb{P}}\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log\left(\frac{p-1}{p}\right)=\sum_{p\in\mathbb{P}}\log(p-1)-\sum_{p\in\mathbb{P}}\log(p)$$

How do you get $\Pi_{n=2}^\infty\frac{n-1}{n}=0$? Not obvious to me. — Gregory Grant, Feb 08 '16 at 21:50
Take the log, and use the fact (Euler) that $\sum \frac{1}{p}$ diverges. — André Nicolas, Feb 08 '16 at 21:51
You might want to check this https://en.wikipedia.org/wiki/Proof_of_the_Euler_product_formula_for_the_Riemann_zeta_function. The $\zeta(1)$ diverges (goes to $\infty$) and it seems to be reciprocal of what you try to evaluate, so your product should converge to $0$. — Sil, Feb 08 '16 at 21:52
$\prod_{n=2}^{\infty}\frac{n-1}{n}$ looks something like $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4} \times...$. We can cancel out the top of each fraction with the bottom of the previous fraction in order to see that this is equal to just $\frac{1}{n}$, which approaches 0 when $n$ approaches infinity. — RSpeciel, Feb 08 '16 at 21:52
I think I have found a duplicate here. The answers there use stronger results, but also prove that $\sum_p \frac{1}{p}$ diverges. — Dietrich Burde, Feb 08 '16 at 22:17
@GregoryGrant, I think what the OP meant is $\prod_{n=2}^\infty{n-1\over n}=\lim_{N\to\infty}\prod_{n=2}^N{n-1\over n}=\lim_{N\to\infty}{1\over N}=0$. I.e., it's a "telescoping" product. — Barry Cipra, Feb 08 '16 at 22:20
to prove that $\prod (1-1/p) = 0$ you have to prove that $\sum_p 1/p = \infty$ and the simplest way is with $\ln \zeta(s) = \sum_p \sum_{r=1}^\infty \displaystyle \frac{p^{-sr}}{r}$ — reuns, Feb 09 '16 at 11:39

Dietrich Burde · Accepted Answer · 2016-02-08T22:37:31.657

2

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.

edited Feb 08 '16 at 22:37

answered Feb 08 '16 at 22:07

Dietrich Burde

140,055

How is $\frac{1}{\zeta(s)}=\prod_{p}\frac{p-1}{p}$? – RSpeciel Feb 08 '16 at 22:13
Because of the Euler product for $\zeta(s)$. – Dietrich Burde Feb 08 '16 at 22:15
@DietrichBurde I think Romain has noticed that the power $s$ is missing. – jim Feb 08 '16 at 22:24
1

I think one would usually not say 'converges to 0'. Usually convergence of an infinite product is to a nonzero finite value. See for instance https://en.wikipedia.org/wiki/Infinite_product – paw88789 Feb 08 '16 at 22:36
@paw88789 corrected. – Dietrich Burde Feb 08 '16 at 22:37
to prove that $\lim_{N \to \infty} \prod_{p \le N} (1-1/p) = \lim_{s \to 1^+} 1/\zeta(s) = 0$ you have to rely on the fact that $1/p > 0$ and that $\zeta$ has a (meromorphic) pole there – reuns Feb 09 '16 at 11:42

score 1 · Answer 2 · answered Feb 09 '16 at 11:26

1

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

answered Feb 09 '16 at 11:26

PITTALUGA

834

Product of Primes

2 Answers2