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.
One of the most 'wonderful' discoveries of the Swiss mathematician Leonhard Euler was the 'product'...
$\displaystyle \sum_{n} \frac{1}{n^{s}} = \prod_{p} \frac{1}{1-\frac{1}{p^{s}}}= \zeta(s)$ (1)
Now if s tends to 1, then $\displaystyle \zeta(s)$ tends to infinity... what's the consequence?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Well, we need the function to be equal to $\displaystyle \frac{\pi^2}{6}$
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
$\displaystyle \zeta(2)\in\mathbb{Q}$
But we know $\displaystyle \zeta(2)\notin\mathbb{Q}$
Therefore, we have reached a contradiction. Correct?
Yes it's correct... but what I had in mind is that the 'infinite sum' $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{s}}$ when $\displaystyle s=1$ is the 'armonic sum' $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ which diverges to $\displaystyle + \infty$. If the 'discovery' of Leonhard Euler is true, then the product $\displaystyle \prod _{p} \frac{1}{1-\frac{1}{p}}$ diverges to $\displaystyle + \infty$ and that means that the product $\displaystyle \prod_{p} (1-\frac{1}{p}) $ diverges to 0. Bur none of the terms $\displaystyle 1-\frac{1}{p}$ is zero, so that necessarly there are infinite factors , i.e. there are infinite primes...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$