Mertens's second theorem states that $$\sum_{p \leq x} \frac 1p = \log(\log x) + M + o(1),$$ where the sum runs over primes $p$ and $M$ is the Meissel-Mertens constant. (There are more precise versions).

I would like to know if there is a really simple proof of the weak form $$\sum_{p \leq x} \frac 1p = \log(\log x) + O(1).$$

I know how to prove one inequality: the fact that every integer can be written as the product of a square and squarefree number shows the inequality $$\sum_{n\leq x} \frac 1n \leq \prod_{p \leq x} \left( 1 + \frac 1p\right) \sum_{m \leq x} \frac 1{m^2} \leq \zeta(2) \prod_{p \leq x} \left( 1 + \frac 1p\right).$$ This inequality and the fact that $\forall t \geq 0, 1 + t \leq \exp(t)$ easily show that $$\sum_{p \leq x} \frac 1p \geq \log(\log x) + O(1)$$ but I can't find such an easy proof for the other inequality.

(There's a quite sibylline reference in Alon and Spencer's The Probabilistic Method, section 4.2, claiming that the estimate I want can be proved by "combining (in a clever way) Stirling's formula with Abel summation." It may be a good answer for my question, but I haven't been able to find a more precise explanation, or to reconstruct this proof from these hints...).


1 Answer 1


I think you mean this. We have $$\log\left(N!\right)=\sum_{p^{k}\leq N}\left[\frac{N}{p^{k}}\right]\log\left(p\right)=N\sum_{p\leq N}\frac{\log\left(p\right)}{p}+O\left(\sum_{p\leq N}\log\left(p\right)\right)$$ $$=N\sum_{p\leq N}\frac{\log\left(p\right)}{p}+O\left(N\right)$$ where the last estimation follows by the Chebyshev theorem. By Stirling$$\log\left(N!\right)=N\log\left(N\right)+O\left(N\right)$$ so $$\sum_{p\leq N}\frac{\log\left(p\right)}{p}=\log\left(N\right)+O\left(1\right).\tag{1}$$Then, using the Abel summation, we get $$\sum_{p\leq N}\frac{1}{p}=\sum_{p\leq N}\frac{\log\left(p\right)}{p}\frac{1}{\log\left(p\right)}$$ $$=\frac{1}{\log\left(N\right)}\sum_{p\leq N}\frac{\log\left(p\right)}{p}+\int_{2}^{N}\sum_{p\leq t}\frac{\log\left(p\right)}{p}\frac{1}{t\log^{2}\left(t\right)}dt$$ and so, by $(1)$, we get $$\sum_{p\leq N}\frac{1}{p}=\log\left(\log\left(N\right)\right)+O\left(1\right)$$ as wanted.

  • $\begingroup$ with slightly more care, I think this method does actually give $o(1)$ $\endgroup$ Oct 26, 2022 at 8:58

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