For $\sigma>1$, put $F(s)=-\zeta'(s)/\zeta(s)$. Its absolutely convergent <Dirichlet series> and
$$
3+4\cos\theta+\cos2\theta=2(1+\cos\theta)^2\geq0
$$
give the <three-four-one zero-free-region argument>
$$
3F(\sigma)+4\Re F(\sigma+it)+\Re F(\sigma+2it)\geq0.
$$
The pole of $\zeta$ at one gives
$$
F(\sigma)=\frac1{\sigma-1}+O(1).
$$

Suppose $\rho=\beta+i\gamma$ is a zero with $\gamma\geq4$ 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 $\rho$ gives
$$
\Re F(\sigma+i\gamma)
\leq-\frac1{\sigma-\beta}+O(\log\gamma).
$$
At $\sigma+2i\gamma$ the same expansion gives merely $\Re F(\sigma+2i\gamma)\leq O(\log\gamma)$. The zero is included in the supplied disk whenever $1-\beta$ and $\sigma-1$ are sufficiently small. Hence
$$
0\leq\frac3{\sigma-1}-\frac4{\sigma-\beta}+O(\log\gamma).
$$

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 give
$$
0\leq\frac3a-\frac4{a+(1-\beta)L}+O(1)<0,
$$
a contradiction. Conjugation handles negative $\gamma$. Reducing $c$ to absorb the bounded range proves the classical <Zero-free region of the Riemann zeta function>
$$
\boxed{\zeta(s)\ne0\quad\text{for}\quad
\sigma\geq1-\frac c{\log|t|},\quad |t|\geq4}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 3 / b / Solution

{c}

Uniformly in a fixed bounded-width disk around height $t$,
$$
\frac{\zeta'(s)}{\zeta(s)}
=\sum_{\rho\text{ nearby}}\frac1{s-\rho}
+O(\log(|t|+3)).
$$
This follows by taking the logarithmic derivative of the Hadamard product for the completed zeta function and estimating the distant zeros.


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 Incoming links: Local partial-fraction expansion of the Riemann zeta logarithmic derivative