Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1

The Von Mangoldt divisor identity givesBy the Stirling formula, the left side is $\log (\lfloor x\rfloor !)=x\log x-x+O(\log x)$, while the assumed Chebyshev estimate gives $\psi (x)=O(x)$. HenceThe contribution of proper prime powers is bounded uniformly:Removing it leaves

PutThe Abel summation formula with weight $1/\log t$ givesSubstituting $A(t)=\log t+E(t)$ yieldsfor a constant $c$. Both final terms are $O(1/\log x)$, soThis is the Mertens second theorem.

The same divisor-identity argument as in part (a) givesOn the other hand, the Abel summation formula givesUnder the proposed asymptotic, the right side isDividing first by $\log x$ gives $a=1$. Subtracting $\log x$, dividing by $\log \log x$, and taking the limit then gives $b=0$. Therefore

Taking the natural logarithm of the finite Euler product and using the Taylor seriesgivesThe double series converges absolutely, and the Mertens second theorem therefore makes the right sideforExponentiating provesThis is the Mertens third theorem.

The prime-factor formula for the Euler totient function isLetThe factors with $p\le y$ contribute at mostby the Mertens third theorem. For $p>y$, the number of distinct prime divisors of $n$ is at most $\log n/\log y$, and henceThus the large-prime product is $1+o(1)$, andTaking reciprocals gives, uniformly as $n\to \infty$,