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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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    That is pretty clear from the integral representation of the $\zeta$ function, or just from Euler's product $$\zeta(s)=\prod_{p\in\mathcal{P}}\left(1-\frac{1}{p^s}\right)^{-1}$$ – 2017-01-29
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    Possible duplicate of [The Riemann zeta function $\zeta(s)$ has no zeros for $\Re(s)>1$](http://math.stackexchange.com/questions/1097963/the-riemann-zeta-function-zetas-has-no-zeros-for-res1) – 2017-01-29
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    @DietrichBurde It seems like that question specifically ask for a proof using the Euler product. – 2017-01-29

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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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    This is the simplest proof I know. (At least, if $\zeta$ is defined as series.) – 2017-01-29
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    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 ? – 2017-01-29
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    It converges absolutely for $\Re(s) > 1$ by the triangle inequality and the fact that $|\mu(n)| \leq 1$. No Euler product needed. – 2017-01-29
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    @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) – 2017-01-29
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    @user1952009 not exactly the same. The difference is one does not need convergence theorems for infinite products to understand this approach. – 2017-01-29
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    Yes, multiplicativity must be used in some way, either in terms of the Euler product or in terms of that showing $f * g = 1_{n = 1}$ and $|\mu(n)| \leq 1$. – 2017-01-29
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    @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. – 2017-01-29
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    So suggesting you can avoid the Euler product here is a very bad idea. – 2017-01-29
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    I don't think avoiding Euler products is necessarily a bad idea. Zero-free regions tend to only really be interesting for Dirichlet series arising from multiplicative functions, and this proof works in that setting. For example, if you want to prove $L(s,\pi)$ is nonvanishing in $\Re(s) > 1$, where $\pi$ is a cuspidal automorphic representation of $\mathrm{GL}_n$, then this proof works. On the other hand, the proof isn't really shorter as the multiplicative inverse of the coefficients of the Dirichlet series $L(s,\pi)$ are more complicated (yet still don't grow too fast). – 2017-01-29
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    @PeterHumphries Come on, you are not serious. So as formal Dirichlet series : $\prod_p (1+\sum_{k \ge 1} p^{-sk}) = \sum_{n=1}^\infty n^{-s}$ so that $\sum_{n=1}^\infty \mu(n) n^{-s} = \prod_p (1-p^{-s}) \implies \mu(n) \in \{-1,0,1\}$ and what Barto wrote follows. That's all I said : don't think you can avoid the Euler product. – 2017-01-29
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    Yes, but that's as formal Dirichlet series. It takes half a page of work to rigorously show that Dirichlet series in their regions of absolute convergence are equal to their Euler product. So either way, you need to do some extra work. – 2017-01-29
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    @PeterHumphries Not at all !! It is obvious that $\prod_{p < N} \frac{1}{1-p^{-s}} = \sum_{n \in A_N} n^{-s}$ where $A_N$ is the set of integers having no prime divisor $\ge N$, and hence $\zeta(s) = \lim_{N \to \infty}\sum_{n \in A_N} n^{-s} = \prod_p \frac{1}{1-p^{-s}}$ for $Re(s) > 1$ where changing the order of summation is allowed – 2017-01-29
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    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? :) – 2017-01-29
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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.