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

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

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

• "A converging infinite product is zero only if one of the factors is zero." But $\prod_{n=1}^\infty{\frac{1}{2}} = 0$. – Dan Brumleve Dec 26 '12 at 11:33
• The statement must be incorrect as Dan points out. – jack Dec 26 '12 at 11:43
• Some people really rush out to downvote answers, which may be wrong but don't deserve to be tossed away so quickly. In this case the answer is correct, though perhaps it could have been better to add a little more explanation. By definition, an infinite CONVERGENT product cannot be zero...that's true! Dan's example is an infinite NON-CONVERGENT product, that's all. I'm upvoting this answer hoping there will be some added explanation, and hopefully whoever downvoted it will think twice about this. – DonAntonio Dec 26 '12 at 12:02
• @DonAntonio, thanks for correcting me, I didn't know that convergent infinite products were defined to exclude zero. I was the one who downvoted. But I still don't see how this answers the question, given that it is just a matter of the definition of convergence. – Dan Brumleve Dec 26 '12 at 12:12
• So ... Dan and everyone else reading this should learn the definition for "convergent" infinite product. And not downvote the next guy who uses the definition correctly. By the way: a convergent infinite product can have value zero, but only if one of the factors is zero. – GEdgar Dec 26 '12 at 14:40

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 $$|\zeta(s)|\ge\frac{3}{4}-\sum_{n\ge 3}\left(\frac{1}{n-1}-\frac{1}{n}\right)=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}.$$

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 $\Re s > 1$.

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 $t \neq 0$.