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

4 Answers 4

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

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    @DanBrumleve, you may feel more comfortable thinking about $\log |\zeta(s)|$ converges for \Re s > 1. (Taking $log$ shows you why 0 should be excluded in the definition of convergent infinite product.)2013-01-08
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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 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}. $

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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 $\Re s > 1$.

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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 $t \neq 0$.