An attempt of proof of Riemann Hypothesis

Abstract

This paper deals with an attempt of proof of the Riemann Hypothesis (RH). Let $T>10^{10}$ arbitrarily large. Let the region $\Omega_T=\Big\{z=x+i y\ \Big|\ \frac{1}{2} 0$. There exists at least one root $\rho=\frac{1}{2}+{\bf u}+i\gamma $ whose real part is greater or equal to the real part of all the other roots in $\Omega_T$. Let $v\geq \frac{3}{2}$. Let $\varepsilon>0$ arbitrarily small. We prove that $f(z)=\frac{\zeta'(z)}{\zeta(z)}$ is analytic in the open disk $\Omega_\varepsilon=\Big\{ \Big|z-\Big(\rho+\frac{\varepsilon}{2}+v\Big)\Big|\Big\}< v.$ Let $s=\rho+\varepsilon$. We prove, from the Taylor series of $\zeta(s)$, that $f(s)\sim \frac{1}{\varepsilon}\rightarrow \infty$ when $\varepsilon\rightarrow 0$, and that, through the representation of $f(s)$ as a Taylor series, $f(s)=f(c_0)-(v-\frac{\varepsilon}{2})f'(c_0) +\frac{(v-\frac{\varepsilon}{2})^2}{2!}f''(c_0)-\frac{(v-\frac{\varepsilon}{2})^3}{3!}f^{(3)}(c_0)+\dots\mbox{\ for\ }c_0=\rho+\frac{\varepsilon}{2}+v,$ in $\Omega_\varepsilon$, that $f(s)\not\rightarrow \infty$ when $\varepsilon\rightarrow 0$, a contradiction which allows us to prove RH.