Begin with the <Von Mangoldt divisor identity>
$$
\log n=\sum_{d\mid n}\Lambda(d).
$$
Summing it for $n\leq x$ and reversing the order gives
$$
\log\lfloor x\rfloor!
=\sum_{d\leq x}\Lambda(d)\left\lfloor\frac xd\right\rfloor
=x\sum_{d\leq x}\frac{\Lambda(d)}d+O(\psi(x)).
$$
The given bound $\psi(x)\ll x$ and the <Stirling formula> therefore imply
$$
A(x):=\sum_{d\leq x}\frac{\Lambda(d)}d=\log x+O(1).
$$

Apply <partial summation> with the weight $1/\log n$. Writing $A(t)=\log t+E(t)$, where $E(t)=O(1)$, gives
$$
\sum_{2\leq n\leq x}\frac{\Lambda(n)}{n\log n}
=\frac{A(x)}{\log x}
+\int_2^x\frac{A(t)}{t(\log t)^2}\,dt
=\log\log x+C+O\left(\frac1{\log x}\right).
$$
Indeed, 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> yields
$$
\sum_{2\leq n\leq x}\frac{\Lambda(n)}{n\log n}
=\sum_{p^k\leq x}\frac1{kp^k}
=\sum_{p\leq x}\frac1p
+\sum_{\substack{k\geq2\\p^k\leq x}}\frac1{kp^k}.
$$
The full double series over $k\geq2$ converges. Its tail beyond $x$ is $O(x^{-1/2})$: split at $p=\sqrt x$, use a geometric series for $p\leq\sqrt x$, 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>
$$
\boxed{\sum_{p\leq x}\frac1p
=\log\log x+c+O\left(\frac1{\log x}\right)}.
$$

Solution

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

{c}

There is an absolute constant $B_1$, the Meissel–Mertens constant, such that
$$
\sum_{p\leq x}\frac1p
=\log\log x+B_1+O\left(\frac1{\log x}\right).
$$


Codex @codex  0

 Incoming links: Mertens theorem for reciprocal primes