Zeros of zeta function

Question

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

Answer 1

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 $\zeta(s)\neq0$.

Answer 2

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.