Infinitely many primes (Riemann Zeta)

**dwsmith** · June 21st 2011, 07:19 PM

Does any one know of a link to the proof that there are infinitely many primes using the Riemann Zeta function or would be willing to produce it?

Thanks.

**chisigma** · June 21st 2011, 08:54 PM

One of the most 'wonderful' discoveries of the Swiss mathematician Leonhard Euler was the 'product'...

$\sum_{n} \frac{1}{n^{s}} = \prod_{p} \frac{1}{1-\frac{1}{p^{s}}}= \zeta(s)$ (1)

Now if s tends to 1, then $\zeta(s)$ tends to infinity... what's the consequence?...

Kind regards

$\chi$ $\sigma$

**dwsmith** · June 22nd 2011, 09:33 AM

Originally Posted by chisigma

One of the most 'wonderful' discoveries of the Swiss mathematician Leonhard Euler was the 'product'...

$\sum_{n} \frac{1}{n^{s}} = \prod_{p} \frac{1}{1-\frac{1}{p^{s}}}= \zeta(s)$ (1)

Now if s tends to 1, then $\zeta(s)$ tends to infinity... what's the consequence?...

Kind regards

$\chi$ $\sigma$

Well, we need the function to be equal to $\frac{\pi^2}{6}$

And if the Riemann Zeta function goes to infinity, then the sum and product are divergent. So we need s = 2.

Assume there are finitely many primes.

Then

$\zeta(2)\in\mathbb{Q}$

But we know $\zeta(2)\notin\mathbb{Q}$

Therefore, we have reached a contradiction. Correct?

**chisigma** · June 22nd 2011, 11:46 AM

Originally Posted by dwsmith

Well, we need the function to be equal to $\frac{\pi^2}{6}$

And if the Riemann Zeta function goes to infinity, then the sum and product are divergent. So we need s = 2.

Assume there are finitely many primes.

Then

$\zeta(2)\in\mathbb{Q}$

But we know $\zeta(2)\notin\mathbb{Q}$

Therefore, we have reached a contradiction. Correct?

Yes it's correct... but what I had in mind is that the 'infinite sum' $\sum_{n=1}^{\infty} \frac{1}{n^{s}}$ when $s=1$ is the 'armonic sum' $\sum_{n=1}^{\infty} \frac{1}{n}$ which diverges to $+ \infty$ . If the 'discovery' of Leonhard Euler is true, then the product $\prod _{p} \frac{1}{1-\frac{1}{p}}$ diverges to $+ \infty$ and that means that the product $\prod_{p} (1-\frac{1}{p})$ diverges to 0. Bur none of the terms $1-\frac{1}{p}$ is zero, so that necessarly there are infinite factors , i.e. there are infinite primes...

Kind regards

$\chi$ $\sigma$

**dwsmith** · June 22nd 2011, 07:43 PM

What is the product of $\zeta(1)$ since the the sum is divergent?

**chisigma** · June 22nd 2011, 08:50 PM

Even if with non precise speaking is $\prod_{p} (1-\frac{1}{p})=\frac{1}{\zeta(1)}=0$ ...

Kind regards

$\chi$ $\sigma$

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