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.

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    $\begingroup$ 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 '17 at 14:04
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    $\begingroup$ @GerhardPaseman: why not posting this as an answer? $\endgroup$ – Seva May 13 '17 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 '17 at 23:08

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.


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