Shrink the constant $c$ from part b if necessary. Put $L=\log|t|$. Zeros in the disk appearing in the supplied partial-fraction formula have $\log|\gamma|=L+O(1/|t|)$, so part b ensures
$$
\beta\leq1-\frac cL
$$
for every such zero.

If $\sigma\geq1+c/(2L)$, absolute convergence of the <logarithmic derivative> gives
$$
\left|\frac{\zeta'(s)}{\zeta(s)}\right|
\leq\sum_{n\geq1}\frac{\Lambda(n)}{n^\sigma}
=-\frac{\zeta'(\sigma)}{\zeta(\sigma)}
\ll\frac1{\sigma-1}\ll L,
$$
with the region $\sigma\geq2$ even easier.

It remains to take $1-c/(2L)<\sigma<1+c/(2L)$. Set
$$
s_0=1+\frac c{2L}+it.
$$
For every local zero, both $\Re(s-\rho)$ and $\Re(s_0-\rho)$ are positive and comparable, while $|s-s_0|\ll1/L$. The partial-fraction formula at $s_0$, together with the preceding Euler-product bound, gives
$$
\sum_\rho\Re\frac1{s_0-\rho}\ll L.
$$
Since $\Re(s_0-\rho)\gg1/L$, it follows that
$$
\sum_\rho\frac1{|s_0-\rho|^2}\ll L^2.
$$
Subtracting the partial-fraction formulas at $s$ and $s_0$ now yields
$$
\left|\frac{\zeta'(s)}{\zeta(s)}-\frac{\zeta'(s_0)}{\zeta(s_0)}\right|
\ll |s-s_0|\sum_\rho\frac1{|s-\rho||s_0-\rho|}+L
\ll L.
$$
Therefore the <logarithmic derivative inside the zeta zero-free region> satisfies
$$
\boxed{\frac{\zeta'(s)}{\zeta(s)}\ll\log|t|}
$$
throughout the required half-width region.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-150/3/c/solution