Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 25 / 2 / a / Solution

Absolute convergence and the Fundamental theorem of arithmetic give the Euler productFor a finite set of primes, expand the geometric factors: their product sums $n^{-s}$ over integers whose prime factors lie in that set. Let the finite sets increase through all primes. Absolute convergence permits passage to the limit and recovers the full Dirichlet series. Moreover ${\sum }_{p,m\ge 1}|p^{-ms}|/m<\infty$, so the logarithm converges and the product has no zeros there.

The same absolutely convergent logarithm, and $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$, giveThis is the product version of the three-four-one zero-free-region argument. It also proves there are no zeros on $\Re s=1$: if $\zeta (1+i{t}_0)=0$ for $t_0\ne 0$, its factor has order at least four as $\sigma \downarrow 1$, while the real pole contributes only order minus three and the $2t_0$ factor remains bounded. The displayed left side would tend to zero, a contradiction. At $t_0=0$ there is a pole, not a zero.

For large $|t|$, put $L=\log (|t|+2)$. The Hardy-Littlewood approximation to the Riemann zeta function at $x\asymp |t|$ gives $|\zeta (\sigma +it)|\ll L$ for $1-2/L\le \sigma \le 3$; its finite sum is bounded by ${\sum }_{n\le x}{n}^{-1+2/L}\ll L$ and the integral term is bounded. The Cauchy estimate for derivatives on circles of radius comparable to $1/L$ consequently gives $|{\zeta }^{\prime }(\sigma +it)|\ll L^2$ for $1-L^{-9}\le \sigma \le 2$.

Take ${\sigma }_0=1+a{L}^{-9}$ with a small fixed $a>0$. The product inequality, $\zeta ({\sigma }_0)\ll L^9/a$, and $|\zeta ({\sigma }_0+2it)|\ll L$ implyIf $1-c{L}^{-9}\le \sigma \le {\sigma }_0$, integration of the derivative along the horizontal segment changes this value by at most $C(a+c)L^{-7}$. Choose $a$ sufficiently small that $Ca<c_0{a}^{3/4}/4$, then $c\le a$ sufficiently small. The lower bound remains a positive multiple of $L^{-7}$. For ${\sigma }_0\le \sigma \le 2$, the same product inequality, $\zeta (\sigma )\le \zeta ({\sigma }_0)$, and the near-one upper bound give that lower bound directly. For $\sigma \ge 2$, the reciprocal Euler product gives $|1/\zeta (s)|\le \zeta (2)$.

Finally the no-zero result on $\Re s=1$, compactness at bounded heights and the regular reciprocal at the pole allow a further fixed reduction of $c$ to include bounded $t$. We have proved the weak logarithmic zero-free region for the Riemann zeta functionThe reciprocal at $s=1$ is its holomorphic extension, equal to zero.