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Well it take my a little bit of time to find it but now I think that my conjecture is ready . Conjecture :

Let $n\geq 100$ and then define the sum :

$$S(n)=\sum_{k=1}^{n}\frac{1}{\operatorname{argtanh}\left(\frac{k}{k+1}\right)}$$

Then we have :

$$\lceil{S(n)}\rceil\leq 2\pi(n)$$

Where we see the ceiling function and the prime counting function .Nicely the equality case is $n=100$ . I have checked my conjecture with wolfram alpha up to $n=1000$ and extra values of $n$ up to $n=1000000$

Obviously I'm inspired by the prime number theorem and it's not a chance if I choose $\operatorname{argtanh}$ in my conjecture .

I have several questions:

Well first I would like to know if it's true for larger $n$ . Is my conjecture equivalent to other conjecture (stronger or weaker conjecture) . Can we improve the conjecture by adding (by example) a fixed exponent ? What my conjecture involves ? If it's not true for small $n$ can we take a larger $n$ to start with ?

Thanks!

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    $\begingroup$ $\text{argtanh} = \tanh^{-1}$? Just making sure, $\endgroup$ Commented Dec 16, 2020 at 15:36
  • $\begingroup$ @JoshuaWang Yes without any doubts ! $\endgroup$ Commented Dec 17, 2020 at 14:28

1 Answer 1

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$$\tanh(x)= \frac{e^{2x}-1}{e^{2x}+1}, \qquad e^{2x}=\frac{1+\tanh(x)}{1-\tanh(x)}$$ $$x = \frac12 \log(\frac{1+\tanh(x)}{1-\tanh(x)}), \quad tanh^{-1}(y)=\frac12 \log(\frac{1+y}{1-y}), \quad tanh^{-1}(1-t)=\frac12 \log(\frac{2-t}{t})$$ Thus for $t$ small enough $$tanh^{-1}(1-t)\ge \frac12\log(1/t)+ \frac14$$

$$\sum_{k=1}^n \frac1{tanh^{-1}(1-\frac1{k+1})}\le O(1)+\sum_{k=1}^n \frac1{\frac12\log(k+1)+\frac14}$$ $$\le O(1)+ 2\sum_{k=1}^n \frac1{\log(k)} (1-\frac14\frac1{\log(k)})\le O(1)+2 Li(n)-\frac18 \frac{n}{\log^2 n}$$

On the other hand the PNT says that $$\pi(n) = Li(n)+O(\frac{n}{\log^3 n})$$ thus for $n$ large enough $$2\pi(n)\ge 1+\sum_{k=1}^n \frac1{tanh^{-1}(1-\frac1{k+1})}$$

It remains to make all the $O$ constants effective and to do a numerical check for $n\le N$.

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