I wondered if it is possible to get a similar inequality like $(1.1)$ of Michael D. Hirschhorn, *Approximating Euler's Constant*, The Fibonacci Quarterly, Volume 49, Number 3 (August 2011) for the Meissel-Mertens constant (as reference I add the Wikipedia link [Meissel–Mertens constant](https://en.wikipedia.org/wiki/Meissel%E2%80%93Mertens_constant) or Theorem 4.12 from [1]).



> 
> **Question.** What can be a good proposal of inequality :
>  $$\text{lower bound}<\sum\_{p\leq n}\frac{1}{p}-\log\log(n)-M<\text{upper bound},$$
>  where $M$ denotes the Meissel–Mertens constant? Or, is it possible to be better with
>  $$\text{lower bound}<\sum\_{p\leq n}\frac{1}{p}-f(n)-M<\text{upper bound}$$
>  for a suitable function $f(n)$?
>  **Many thanks.**
> 
> 
> 


If it is already in the literature, add the reference and I would try to search and read the inequality from the source, or add yourself inequality or improvements.


References:
-----------


[1] Tom M. Apostol, *Introduction to Analytic Number Theory*, Springer (1976).



Sharp estimates for Meissel-Mertens constant
Already Mertens has proved in 1874 (*Ein Beitrag zur analytyischen Zahlentheorie*, J. Reine Angew. Math. 78) a refined variant of the result, which can be found in the book of Apostol. He showed that
$$\tag{1}\sum\_{p \le x} \frac{1}{p} = \log \log x +M + \mathcal{O}\Big(\frac{1}{\log{x}}\Big).$$
A proof can be found in the book 'Multiplicative Number Theory I. Classical Theory' by Montgomery and Vaughan, see Theorem 2.7 (d). Since this is a standard fact, it can be found in most books on analytic number theory. Moreover, there is an interesting paper on Mertens proof, written by Villarino: See [Mertens' Proof of Mertens' Theorem](https://www.researchgate.net/publication/2119159_Mertens%27_Proof_of_Mertens%27_Theorem).


In particular, (1) can be strengthened. Mertens has gave an explicit constant for the $\mathcal{O}$-term. (The error is at most $4/\log{x}$.) But this is not the best known-bound, see [here](https://en.wikipedia.org/wiki/Mertens%27_theorems#Mertens'_second_theorem_and_the_prime_number_theorem). It is easy to see (using partial summation) that the error bound (1) is directly related to the prime number theorem. Wikipedia says that the strongest form known of the prime number theorem implies that
$$\sum \_{p\leq x}{\frac {1}{p}=\log \log x+M+O(e^{-c(\log x)^{3/5}(\log \log x)^{-1/5}})}$$


for some $c > 0$. 


