Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 150 / 1

Begin with the Von Mangoldt divisor identitySumming it for $n\le x$ and reversing the order givesThe given bound $\psi (x)\ll x$ and the Stirling formula therefore imply

Apply partial summation with the weight $1/\log n$. Writing $A(t)=\log t+E(t)$, where $E(t)=O(1)$, givesIndeed, the integral of $E(t)/(t(\log t{)}^2)$ converges, and its tail from $x$ to infinity is $O(1/\log x)$.

Grouping the left side by prime powers yieldsThe full double series over $k\ge 2$ converges. Its tail beyond $x$ is $O(x^{-1/2})$: split at $p=\sqrt{x}$, use a geometric series for $p\le \sqrt{x}$, and compare ${\sum }_{p>\sqrt{x}}{p}^{-2}$ with the corresponding sum over integers. Absorbing its limit into the constant proves the Mertens theorem for reciprocal primes

Putby part a. Double-counting divisibility gives the first moment of the prime omega function:Moreover,soExpanding the square and using the given bound $\pi (x)\ll x/\log x$ now gives the Turán normal-order theorem for distinct prime divisors estimate

By the Chebyshev inequality, the number of $n\le x$ for whichis $O(x/L^{1/2})=o(x)$. Discard the $O(\sqrt{x})$ integers below $\sqrt{x}$. For $\sqrt{x}<n\le x$,and $(\log \log n{)}^{3/4}\sim L^{3/4}$. Hence, for all sufficiently large $x$, every remaining integer counted in the question also satisfies the preceding inequality. Therefore