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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 $$\zeta(s)=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1}$$ $\endgroup$ Jan 29 '17 at 16:15
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    $\begingroup$ Possible duplicate of The Riemann zeta function $\zeta(s)$ has no zeros for $\Re(s)>1$ $\endgroup$ Jan 29 '17 at 16:20
  • $\begingroup$ @DietrichBurde It seems like that question specifically ask for a proof using the Euler product. $\endgroup$ Jan 29 '17 at 17:00
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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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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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    $\begingroup$ 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
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    $\begingroup$ @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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    $\begingroup$ @PeterHumphries What I mean witth "something important" is that for any Dirichlet series $F(s) = \sum_{n=1}^\infty a_n n^{-s}$, finding a bound for the coefficients $b_n$ of $\frac{1}{F(s)} = \sum_{n=1}^\infty b_n n^{-s}$ is much harder whenever $a_n$ isn't multiplicative. $\endgroup$
    – reuns
    Jan 29 '17 at 17:55
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    $\begingroup$ So suggesting you can avoid the Euler product here is a very bad idea. $\endgroup$
    – reuns
    Jan 29 '17 at 17:58
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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? :) $\endgroup$ Jan 29 '17 at 19:55
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Here is a more direct proof that $\zeta(s) \neq 0$ for $s \in \mathbb{C}$ with $\text{Re }s>1$. It doesn't use any complex analysis or Dirichlet series. It only uses the Euler product and some basic knowledge about sequences and series.

Let $p_1,p_2,...$ be the sequence of all the primes, i.e. $p_1=2$, $p_2=3$, $p_3=5,...$. The finite products $$\prod_{n=1}^k \frac{1}{1-p_n^{-s}}$$ are always $\neq 0$. We can therefore write

$$|\zeta(s)|=\biggr|\prod_{n=1}^{\infty}\frac{1}{1-p_n^{-s}}\biggl|=\prod_{n=1}^{\infty}\biggr|\frac{1}{1-p_n^{-s}}\biggl|=\lim_{k \to \infty}\prod_{n=1}^{k}\biggr|\frac{1}{1-p_n^{-s}}\biggl|$$

$$=\lim_{k\to \infty}\exp\biggr(\log\prod_{n=1}^{k}\biggr|\frac{1}{1-p_n^{-s}}\biggl|\biggl)=\lim_{k\to \infty}\exp\biggr(\sum^{k}_{n=1}\log\biggr|\frac{1}{1-p_n^{-s}}\biggl|\biggl)$$

$$=\lim_{k\to \infty}\exp\biggr(-\sum^{k}_{n=1}\log|1-p_n^{-s}|\biggl)$$

It is sufficient to show that the sum in the last line converges for $k \to \infty$. The result then follows because $\exp$ is continuous and never $0$, meaning that $$\lim_{k \to \infty} \exp(x_k)=\exp(\lim_{k \to \infty} x_k)\neq 0$$ for every convergent sequence $(x_k)$. In order to show convergence, we can show absolute convergence using the comparison test: $$\biggr|\log|1-p_n^{-s}|\biggl| \leq \biggr|\log(1+|p_n^{-s}|)\biggl|= \log(1+|p_n^{-s}|)\leq |p_n^{-s}|=p_n^{-\sigma}$$ where $\sigma=\text{Re }s$. Now, the result follows from $\sigma >1$ and $$\sum_{n=1}^{\infty} p_n^{-\sigma}\leq \sum_{n=1}^{\infty}n^{-\sigma}$$

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