Question about Riemann $\zeta(s)$ function zeroes How can it be shown that the Riemann $\zeta(s)$ function has no zeroes for $\Re(s) > 1$?
 A: For $\sigma>1$, we have the converging Euler product. A converging infinite product is zero only if one of the factors is zero.
A: 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
$$
|\zeta(s)|\ge\frac{3}{4}-\sum_{n\ge 3}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}.
$$
A: Convergence of the Euler product breaks down to the following independent statements:


*

*$\displaystyle \prod_{p \le N} (1-p^{-s})^{-1} \to \zeta(s)$ 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.

*$\displaystyle \sum_p |\ln (1-p^{-s})| < +\infty$ 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 $\Re s > 1$.
A: I present my proof which employs Riemann's integral property:
Proof.
For $\alpha>1$ and $t=0$ we get $\sum_{k=1}^{\infty}
\frac{1}{k^s}=\sum_{k=1}^{\infty} \frac{1}{k^{\alpha}}>0$.
For $s=\alpha-it(\alpha>1, t \in R \setminus \{0\})$, we set
\begin{equation*}
S_n = \sum_{k=1}^n \frac{1}{k^s}= \sum_{k=1}^n
\frac{1}{k^{\alpha}}e^{it\ln k}=(S^{(1)}_n , S^{(2)}_n ),
\end{equation*}
\begin{equation}
S_{n+1} = \sum_{k=1}^{n+1} \frac{1}{k^s}= \sum_{k=1}^{n+1}
\frac{1}{k^{\alpha}}e^{it\ln k}=(S^{(1)}_{n+1} , S^{(2)}_{n+1} ).
\end{equation}
Let us consider a line $l_n$ on the complex plane, defined by the
points $S_n$ and $S_{n+1}$. This line is described by the
following equation
\begin{equation}
 \frac{ x - S^{(1)}_ n}{ y - S^{(2)}_n} = \frac{S^{(1)}_{n+1} - S^{(1)}_n} { S^{(2)}_{n+1} -
 S^{(2)}_n }.
 \end{equation}
The normal form of this equation is
\begin{equation*}
  x \sin(t \ln(n+1))-y
\cos(t(\ln(n+1)))+ \sum_{k=1}^{n+1}\frac{1}{k^{\alpha}}\sin(t \ln
k) \cos(t\ln(n+1))-
\end{equation*}
\begin{equation}
\cos(t \ln k) \sin(t\ln(n + 1)) = 0.
\end{equation}
Hence, the distance $\rho(0, l_n)$  between the origin $(0, 0)$ of the
plane and the line $l_n$ is calculated by
\begin{equation*}
\rho(0, l_n) = \big| \sum_{k=1}^{n+1}\frac{1}{k^{\alpha}}(\sin(t
\ln k) \cos(t \ln(n+1))-\cos(t \ln k) \sin(t \ln(n+1))\big| =
\end{equation*}
\begin{equation}
\big| \sum_{k=1}^{n+1}\frac{1}{k^{\alpha}} (\sin(t \ln(k) -t
\ln(n+1))\big| =\big| \sum_{k=1}^{n+1}\frac{1}{k^{\alpha}}(\sin(t
\ln(\frac{k}{n+1}))\big|.
\end{equation}
We have
 \begin{equation*}
  \lim_{n \to \infty} |S_n| \ge \lim_{n \to \infty} \rho(0, l_n) =\lim_{n \to \infty}\big| \sum_{k=1}^{n+1}\frac{1}{k^{\alpha}}\sin(t \ln
(\frac{k}{n+1}))\big|=
\end{equation*}
\begin{equation*}
\lim_{n \to \infty} \frac{1}{{(n+1)}^{\alpha}}\big|
\sum_{k=1}^{n+1}\frac{1}{{(\frac{k}{n+1}}^{\alpha})}\sin(t \ln
(\frac{k}{n+1}))\big|=
\end{equation*}
\begin{equation*}
\lim_{n \to \infty} \frac{n+1}{{(n+1)}^{\alpha}}\big|\frac{1}{n+1}
\sum_{k=1}^{n+1}\frac{1}{{(\frac{k}{n+1})}^{\alpha}}\sin(t \ln
(\frac{k}{n+1}))\big|\ge
\end{equation*}
\begin{equation*}
\lim_{n \to \infty} \big|\frac{1}{n+1}
\sum_{k=1}^{n+1}\frac{1}{{(\frac{k}{n+1})}^{\alpha}}\sin(t \ln
(\frac{k}{n+1}))\big|=
\end{equation*}
\begin{equation*}
\big | \int_0^1 \frac{\sin(t\ln x)}{x^{\alpha}}dx \big
|=\frac{|t|}{(1-\alpha)^2+t^2}>0,
\end{equation*}
because $t \neq 0$.
