The <Von Mangoldt divisor identity> gives
$$
\sum_{n\leq x}\log n
=\sum_{d\leq x}\Lambda(d)\left\lfloor\frac xd\right\rfloor
=x\sum_{d\leq x}\frac{\Lambda(d)}d+O(\psi(x)).
$$
By the <Stirling formula>, the left side is $\log(\lfloor x\rfloor!)=x\log x-x+O(\log x)$, while the assumed <Chebyshev estimate> gives $\psi(x)=O(x)$. Hence
$$
\sum_{n\leq x}\frac{\Lambda(n)}n=\log x+O(1).
$$
The contribution of proper <prime powers> is bounded uniformly:
$$
\sum_{\substack{p^k\leq x\\k\geq2}}\frac{\log p}{p^k}
\leq\sum_p\frac{\log p}{p(p-1)}<\infty.
$$
Removing it leaves
$$
\boxed{\sum_{p\leq x}\frac{\log p}{p}=\log x+O(1).}
$$

Solution

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

{c}

The Chebyshev estimate is
$$
\psi(x)\asymp x.
$$
It gives upper and lower bounds of the correct linear order without asserting the <Prime number theorem> asymptotic $\psi(x)\sim x$.

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 Incoming links: Chebyshev estimate