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I am a beginner of analytic number theory. In studying about zero of zeta function, i faced a Question.

Is there any zero $s=\sigma +t i$ of Riemann zeta function such that $\sigma> 1$ ??There is no answer, of course.

I want to know its proof as simple as possible

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    $\begingroup$ That is pretty clear from the integral representation of the $\zeta$ function, or just from Euler's product $$\zeta(s)=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1}$$ $\endgroup$ – Jack D'Aurizio Jan 29 '17 at 16:15
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    $\begingroup$ Possible duplicate of The Riemann zeta function $\zeta(s)$ has no zeros for $\Re(s)>1$ $\endgroup$ – Dietrich Burde Jan 29 '17 at 16:20
  • $\begingroup$ @DietrichBurde It seems like that question specifically ask for a proof using the Euler product. $\endgroup$ – punctured dusk Jan 29 '17 at 17:00
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Using Dirichlet series (no complex analysis required)

Theorem. Suppose $f,g$ are arithmetic functions such that $$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}\quad\text{and}\quad G(s)=\sum_{n=1}^\infty\frac{g(n)}{n^s}$$ converge absolutely at $s\in\mathbb C$. Let $h=f*g$ be their Dirichlet convolution. Then $$F(s)G(s)=\sum_{n=1}^\infty\frac{h(s)}{n^s}=: H(s)$$ (and the latter series converges absolutely at $s$).

Proof. Given absolute convergence, we can rearrange the terms. $\square$


With $f=\bf1$ and $g=\mu$ we have $f*g=1$ for $n=1$ and $0$ otherwise.

If $\sigma>1$, both $F(s)=\zeta(s)$ and $G(s)$ converge absolutely. So

$$\zeta(s)G(s)=1.$$

In particular, it follows that $\zeta(s)\neq0$.

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  • $\begingroup$ This is the simplest proof I know. (At least, if $\zeta$ is defined as series.) $\endgroup$ – punctured dusk Jan 29 '17 at 16:59
  • $\begingroup$ I think you missed something important. How do you show that $\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n)n^{-s}$ converges for $Re(s) > 1$ without the Euler product ? $\endgroup$ – reuns Jan 29 '17 at 17:27
  • $\begingroup$ It converges absolutely for $\Re(s) > 1$ by the triangle inequality and the fact that $|\mu(n)| \leq 1$. No Euler product needed. $\endgroup$ – Peter Humphries Jan 29 '17 at 17:49
  • $\begingroup$ @PeterHumphries How do you show $|\mu(n)|\le 1$ without the Euler product ?.. (hint : by using that $1$ is completely multiplicative... which is exactly the same as the Euler product) $\endgroup$ – reuns Jan 29 '17 at 17:50
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    $\begingroup$ Come on, we all agree that there's a little bit of work to be done to go from the series to the Euler product (which may not be obvious to someone who is new to analytic nt), and we all agree that the Euler product is intimately related to multiplicativity. Anything else that has to be said? :) $\endgroup$ – punctured dusk Jan 29 '17 at 19:55
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By Hurwitz's theorem, if $U \subseteq \Bbb{C}$ is a connected open set and $(f_n)$ is a sequence of analytic functions on $U$ which do not attain the value zero on $U$ and converge to an analytic function $f$ uniformly on compact subsets of $U$, then $f$ is either identically zero or never attains the value zero on $U$. The proof is an easy application of the argument principle.

If we take $f_n(s) = \prod_{m=1}^n (1-p_m^{-s})^{-1}$ and $U =\{s \in \Bbb{C}: Re(s)>1\}$, then $(f_n)$ converges uniformly on compact subsets of $U$ to $\zeta(s)$. Since $\zeta(s)$ is not identically zero and none of the $f_n$ attain the value zero on $U$, it follows that $\zeta$ has no zeros in $U$.

It is necessary to use complex analytic techniques to prove this, I believe. Maybe you can get away with a longer argument that uses only real variable techniques somehow.

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