Primes numbers and Riemann zeta function.

Question 1: Is there a proof of the infinitude of prime numbers using the Riemann Zeta function. Exboço could show me a proof of this where I could find it?

Question 2: What is the relationship of this function with the famous Riemann hypothesis?

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    $\begingroup$ For 2, see this and this. $\endgroup$ – J. M. is a poor mathematician Dec 20 '11 at 16:07
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    $\begingroup$ Q2: Which function? If zeta, of course, the RH is about the zeroes of the zeta function. $\endgroup$ – N. S. Dec 20 '11 at 16:08
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    $\begingroup$ $\zeta(1)= \prod_{p {\rm prime}} \frac{1}{1- \frac{1}{p}} = \infty $ This guarantees that there are infinitely many primes... Also PNT is usually proven using properties of the RH..... Last but not least, if there would be finitely many primes then the Euler product would be finite, thus "convergent" everywhere... Which would imply many things known not to be true (for example that $\zeta$ has no zero)... $\endgroup$ – N. S. Dec 20 '11 at 16:12
  • $\begingroup$ For 1, see this. $\endgroup$ – J. M. is a poor mathematician Dec 20 '11 at 16:13
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    $\begingroup$ I'm sorry. I do not search for a tag on "Riemann-function Zenta." There execelentes answers in this tag. $\endgroup$ – MathOverview Dec 20 '11 at 16:30

It is easy to lose track of the Riemann Hypothesis behind all the noise about the Riemann Hypothesis. But following the usual statement of the problem (where I use the Clay Mathematics Institute and their prize as the usual statement), the hypothesis is that all the zeroes of the Riemann Zeta function occur on the line $\Re(s) = 1/2$.

With respect to your first question, the Riemann Zeta Function is traditionally used in one proof of the Prime Number Theorem (which states the asymptotic density of primes). In this way, it gives that there are infinitely many primes.

Alternatively, as pointed out by N.S. in the comments, the fact that $\zeta(1) = \infty$ is a direct statement that there are infinitely many primes. And 'calculating' this is not so bad, as long as you know that the harmonic series diverges and that $\sum\limits \frac{1}{n} = \prod_p \frac{1}{1 - p^{-1}}$.


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