The same divisor-identity argument as in part (a) gives
$$
\sum_{n\leq x}\frac{\Lambda(n)}n=\log x+O(1).
$$
On the other hand, the <Abel summation formula> gives
$$
\sum_{n\leq x}\frac{\Lambda(n)}n
=\frac{\psi(x)}x+\int_1^x\frac{\psi(t)}{t^2}\,dt.
$$
Under the proposed asymptotic, the right side is
$$
a\log x+b\log\log x+o(\log\log x)+O(1).
$$
Dividing first by $\log x$ gives $a=1$. Subtracting $\log x$, dividing by $\log\log x$, and taking the <limit> then gives $b=0$. Therefore
$$
\boxed{a=1,\qquad b=0.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-150/1/c/solution