Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 150 / 3

Let $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converge absolutely for $\Re s>{\sigma }_a$. Perron formula states that for $c>{\sigma }_a$ and nonintegral $x>0$,where the integral is understood as the limit of symmetric truncations under the usual convergence hypotheses. If $x$ is an integer, the endpoint term is counted with weight $1/2$. Effective versions truncate at height $T$ and include an explicit error depending on the coefficients and the distance of $x$ from nearby integers.

Solved by gpt-5.6-sol high.

Put $t=s-1$. The Laurent series of the logarithmic derivative at the simple pole of $\zeta$ has the formwhere in fact $c_0=\gamma$. HenceThe coefficient of $t^{-1}$ in their product, and therefore the residue, iswith the constant $A=-1-2c_0$; equivalently, $A=-1-2\gamma$.

Solved by gpt-5.6-sol high.

The Von Mangoldt function is nonnegative and satisfies $\Lambda (m)\le \log m$. Using the Von Mangoldt divisor identity,

For $0<u\le 1$, a comparison with an improper integral gives

Solved by gpt-5.6-sol high.

For $\Re s>1$, the Dirichlet series multiplication rule and $-{\zeta }^{\prime }/\zeta (s)={\sum }_{n\ge 1}\Lambda (n)n^{-s}$ giveApply an effective Perron formula on the line $\kappa =1+1/\log x$ and truncate atThe bound from part (c) controls the truncation error.

Use the classical Zero-free region of the Riemann zeta functiontogether with ${\zeta }^{\prime }/\zeta (s)\ll {\log }^2(|\Im s|+3)$ there. Contour shifting moves the Perron contour to $\Re s=1-c_1/\log T$. The only crossed singularity is the double pole at $s=1$, whose residue is $x\log x+Ax$ by part (b). On the new contour,and the logarithmic-derivative bounds contribute only powers of $\log x$, which can be absorbed by reducing the positive constant in the exponential. The horizontal integrals and Perron truncation error are $O(x\exp (-c_2\sqrt{\log x}))$ as well. Therefore, for some $c>0$,

Solved by gpt-5.6-sol high.