2
$\begingroup$

Let $p_k$ denote the $k-th$ prime. Do there exist some constant $A>0$ such that for every sufficiently large $k$,

$$\sum_{p\leq p_k} \frac{1}{p} > B + \log\log p_k + \frac{A}{\log p_k}$$

where $B$ is the Mertens constant?

It is known that

$$\sum_{p\leq p_k} \frac{1}{p} = B + \log\log p_k + O\left(\frac{1}{\log p_k} \right)$$

but nothing seems to be known about the sign of the implicit constant in the Landau $O$-symbol.

Improve this question
$\endgroup$
3
  • 5
    $\begingroup$ No. Mertens original paper has the opposite inequality with some terms plus 4/log(n+1), which means you can compute n_0 for which A=5 works for n bigger than n_0 for the opposite inequality. Also, there is oscillation (see Diamond and Pintz) which means your inequality with any positive A fails infinitely often. Gerhard "Goes This Way And That" Paseman, 2017.05.13. $\endgroup$
    – Gerhard Paseman
    May 13, 2017 at 14:04
  • 1
    $\begingroup$ @GerhardPaseman: why not posting this as an answer? $\endgroup$
    – Seva
    May 13, 2017 at 16:11
  • $\begingroup$ Because I don't know if the question has a typo. If the question is meant as asked, I may post an answer. Gerhard "Isn't Sure Of The Question" Paseman, 2017.05.13. $\endgroup$
    – Gerhard Paseman
    May 13, 2017 at 23:08

1 Answer 1

Reset to default
5
$\begingroup$

I recommend https://arxiv.org/abs/math/0504289 on the proof of Mertens by Mark Villarino. He shows that in fact the opposite inequality (less than) holds with A=4, and also shows the error term from the original proof (with an additional 2/(n ln n) term) and calculation of B. I believe there are some oscillation results by Diamond and Pintz which would show that the posted inequality (greater than) never holds for all n sufficiently large for any positive A .

Gerhard "Now On The Right Side" Paseman, 2017.05.14.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.