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

For $\sigma >1$, unique prime factorization and absolute convergence give the Euler productEvery factor is nonzero and the product converges to a nonzero limit. Equivalently, the absolutely convergent identityprovides a reciprocal. This proves the Euler-product nonvanishing of the Riemann zeta function.

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

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 ensuresfor every such zero.

If $\sigma \ge 1+c/(2L)$, absolute convergence of the logarithmic derivative giveswith the region $\sigma \ge 2$ even easier.

It remains to take $1-c/(2L)<\sigma <1+c/(2L)$. SetFor every local zero, both $\Re (s-\rho )$ and $\Re (s_0-\rho )$ are positive and comparable, while $|s-s_0|\ll 1/L$. The partial-fraction formula at $s_0$, together with the preceding Euler-product bound, givesSince $\Re (s_0-\rho )\gg 1/L$, it follows thatSubtracting the partial-fraction formulas at $s$ and $s_0$ now yieldsTherefore the logarithmic derivative inside the zeta zero-free region satisfiesthroughout the required half-width region.