For $\sigma>1$, the <three-four-one zero-free-region argument> starts from
$$
3+4\cos\theta+\cos2\theta=2(1+\cos\theta)^2\geq0.
$$
Applying this termwise to
$$
-\frac{\zeta'(s)}{\zeta(s)}
=\sum_{n\geq1}\frac{\Lambda(n)}{n^s}
$$
gives
$$
-3\frac{\zeta'(\sigma)}{\zeta(\sigma)}
-4\Re\frac{\zeta'(\sigma+it)}{\zeta(\sigma+it)}
-\Re\frac{\zeta'(\sigma+2it)}{\zeta(\sigma+2it)}
\geq0.
$$

Suppose $\rho=\beta+i\gamma$ is a zero with $|\gamma|\geq4$. In the supplied <Local partial-fraction expansion of the Riemann zeta logarithmic derivative>, every nearby zero contributes a nonnegative real part at $\sigma+i\gamma$ when $\sigma>1$. Keeping the term from $\rho$ gives
$$
\Re\frac{\zeta'(\sigma+i\gamma)}{\zeta(\sigma+i\gamma)}
\geq\frac1{\sigma-\beta}-O(\log|\gamma|).
$$
At height zero the pole at one gives
$$
-\frac{\zeta'(\sigma)}{\zeta(\sigma)}
=\frac1{\sigma-1}+O(1),
$$
and the same local expansion at height $2\gamma$ gives
$$
-\Re\frac{\zeta'(\sigma+2i\gamma)}{\zeta(\sigma+2i\gamma)}
\ll\log|\gamma|.
$$
Substitution yields
$$
\frac4{\sigma-\beta}
\leq\frac3{\sigma-1}+O(\log|\gamma|).
$$
Set $\sigma=1+\eta/\log|\gamma|$, first choosing a sufficiently small absolute $\eta>0$. If $\beta>1-c/\log|\gamma|$, the left side is at least $4\log|\gamma|/(\eta+c)$. Choosing $c>0$ sufficiently small contradicts the last inequality. Therefore
$$
\boxed{\zeta(s)\ne0
\quad\text{for}\quad
\sigma>1-\frac c{\log|t|},\quad |t|\geq4.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-150/3/c/solution