Question about Riemann $\zeta(s)$ function zeroes

Question

How can it be shown that the Riemann $\zeta(s)$ function has no zeroes for $\Re(s) > 1$?

Answer 1

For $\sigma>1$, we have the converging Euler product. A converging infinite product is zero only if one of the factors is zero.

Answer 2

Convergence of the Euler product breaks down to the following independent statements:

1. $\displaystyle \prod_{p \le N} (1-p^{-s})^{-1} \to \zeta(s)$ as $N \to \infty$ due to absolute convergence of the series $\displaystyle \zeta(s) = \sum_n n^{-s}$ and the usual "rearrangement of terms" trick.

2. $\displaystyle \sum_p |\ln (1-p^{-s})| < +\infty$ with $\ln$ being the branch of logarithm defined by $\ln 1 = 0$. This follows from $|\ln (1-p^{-s})| \sim |p^{-s}|$ and $\sum_p |p^{-s}| < +\infty$ for $\Re s > 1$.

Answer 3

Edit: Titchmarsh offers the following direct proof that does not require knowledge of infinite products: For fixed $\sigma>1$, we can show there are no zeros with real part $\ge\sigma$ by considering, for a parameter $P$ $$ \prod_{\substack{p\text{ prime}\\p\le P}}\left(1-p^{-s}\right)\zeta(s)=1+m_1^{-s}+m_2^{-s}\ldots $$ where $m_1$, $m_2,\ldots$ are all the integers all of whose prime factors exceed $P$. Thus $$ \left|\prod_{p\text{ prime},p<P}\left(1-p^{-s}\right)\zeta(s)\right|\ge 1-\sum_{n=P+1}^\infty \frac{1}{n^\sigma}\ge 1-\int_{P}^\infty x^{-\sigma}\, dx=1-\frac{P^{1-\sigma}}{\sigma-1}. $$ For fixed $\sigma$ we can find $P$ sufficiently large that the right side is $>0$.

Original Answer: On the other hand, it is elementary that $\zeta(s)$ has no zero for $\sigma$ the real part of $s\ge2$. In this region $$ |\zeta(s)|\ge 1-\sum_{n\ge 2}\frac{1}{n^\sigma}\ge1-\sum_{n\ge 2}\frac{1}{n^2}\ge1-\frac{1}{4}-\sum_{n\ge3}\frac{1}{(n-1)n}. $$ Thus $$ |\zeta(s)|\ge\frac{3}{4}-\sum_{n\ge 3}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}. $$