 Solution

For $\sigma >1$, put $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$. Its absolutely convergent Dirichlet series andgive the three-four-one zero-free-region argumentThe pole of $\zeta$ at one gives

Suppose $\rho =\beta +i\gamma$ is a zero with $\gamma \ge 4$ and $\beta$ close to one. Apply the supplied Local partial-fraction expansion of the Riemann zeta logarithmic derivative at $\sigma +i\gamma$. Every term has positive real part, so retaining the term belonging to $\rho$ givesAt $\sigma +2i\gamma$ the same expansion gives merely $\Re F(\sigma +2i\gamma )\le O(\log \gamma )$. The zero is included in the supplied disk whenever $1-\beta$ and $\sigma -1$ are sufficiently small. Hence

Set $L=\log \gamma$ and $\sigma =1+a/L$, where $a>0$ is a sufficiently small fixed constant. If $(1-\beta )L$ were smaller than a sufficiently small constant $c>0$, division by $L$ would givea contradiction. Conjugation handles negative $\gamma$. Reducing $c$ to absorb the bounded range proves the classical Zero-free region of the Riemann zeta function