# Riemann Zeta - Euler product convergence?

I'm fascinated by the Riemann Zeta hypothesis - I haven't yet taken a course in complex analysis but I'm curious what this sentence means on Wikipedia:

The convergence of the Euler product shows that ζ(s) has no zeros in this region, as none of the factors have zeros.

I'm wondering if a) the convergence of the Euler product comes purely from the fact that the p-series diverges when the exponent is greater than 1? b) What the sentence from Wikipedia is essentially saying - I'm not understanding what factors are being referred to or how the convergence shows that the Zeta function has no zeroes for $\Re(s) > 1$.

Edit: I was looking over this answer and it seems to be similar, but how do we know that, as this author wrote, $(1-p^{-s})^{-1}\neq 0$ for all primes $p$? Can't it converge to 0, meaning that $(1-p^{-s})^{-1} = 0$ for at least one prime $p$?

• The idea is that $\zeta(s) =\prod_p \frac{1}{1-p^{-s}} = \exp(F(s))$ where $F(s) = -\sum_p \log(1-p^{-s}) = \sum_{p^k} \frac{p^{-sk}}{k}$ is analytic on $\Re(s) > 1$, thus $\zeta(s)$ has no zeros there. – reuns Jul 6 '17 at 6:16
• $(1-p^{-s})^{-1}$ is between $1$ and $2$ for every real positive $p$. – N74 Jul 6 '17 at 6:18
• Thanks for the answers. So like here: math.stackexchange.com/a/1097980/268374, why do we know $(1-p^{-s})^{-1}\neq 0$ for all primes $p$? – rb612 Jul 6 '17 at 6:20
• Also note that when $s < 0$ : $\prod_p \frac{1}{1-p^{-s}} = 0$ (the product is said to diverge to $0$). That's why it is important to restrict to $\Re(s) > 1$, where we have the equality $\sum_{n=1}^\infty n^{-s} = \prod_p \frac{1}{1-p^{-s}}$. – reuns Jul 6 '17 at 6:24
• That's what I said : here the product converges and its factor are non-zero for $\Re(s) > 1$ really means that $-\sum_p \log(1-p^{-s}) = \log(\prod_p \frac{1}{1-p^{-s}})=\log \zeta(s)$ is analytic for $\Re(s) > 1$. – reuns Jul 6 '17 at 6:29

$$\zeta(s) = \sum_{n=1}^{\infty} n^{-s}$$
for $$\Re(s)>1$$ is convergent and admits the Euler product representation
$$\zeta(s) = \prod_p (1-p^{-s})^{-1}.$$
Since none of the factors in the product equal zero (i.e., $$p^{-s}$$ is never equal to $$1$$), $$\zeta(s)$$ is never equal to zero in the half-plane $$\Re(s)>1$$.