Mertens' Theorem states that
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + O(1/\log x).$$
This is weaker than the prime number theorem; in fact according to the [Wikipedia page](https://en.wikipedia.org/wiki/Mertens%27_theorems), the prime number theorem is equivalent to
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + o(1/\log x).$$
However no reference or proof is given, it just simply says that "Although this equivalence is not explicitly mentioned there, it can for instance be easily derived from the material in chapter I.3 of: G. Tenenbaum. Introduction to analytic and probabilistic number theory."


So what is a proof/reference for this fact? I naturally tried to prove it myself using partial summation, and it is easy to see where the $\log \log x$ comes from, however the constant $M$ and $o(1/\log x)$ are a bit more mysterious.


Standard proofs in prime number theory often proceed by introducing a logarithmic weight via the Von Mangoldt function, then proving an asymptotic and going back again. I would prefer to avoid such an approach as I have a different problem in mind coming from counting rational points on varieties where I can't introduce a logarithmic weight.



Proving Mertens' theorem using the prime number theorem
Well, one can always say that the PNT is equivalent to
$$\sum\_{p \leq x}\frac{1}{p} = \log \log x + M + o\left(\frac{1}{\log x}\right),\tag{$\ast$}$$
because both results are true (with better error terms). This is of course not what is meant by the Wikipedia page. Instead, the idea is that the equivalence PNT$\,\Leftrightarrow(\ast)$ can be established in a simpler way than either PNT or $(\ast)$. On the other hand, "simpler" is a subjective word, e.g. I usually find Tauberian arguments tricky.


At any rate, the first three exercises in Section 8.1.1 of Montgomery-Vaughan: Multiplicative number theory I address this question. For example, the PNT easily implies the relation $\psi(x)\sim x$ (logarithmic weights!), which then implies rather nontrivially (using Theorem 8.1 = Axer's theorem) that
$$\sum\_{p\leq x}\frac{\log p}{p}=\log x+C+o(1)$$
for some constant $C$. From here, $(\ast)$ follows easily by partial summation.


