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It is well-known that $|\pi(x)-\operatorname{li}(x)| \leq \epsilon(x)$, where $\pi(x) = \sum \limits_{p \leq x} 1$ is the prime counting function, where $\operatorname{li}(x) = \int \limits_{2}^{x}\frac{1}{\ln(t)} dt$ is the logarithmic integral, and where the function $\epsilon$ satisfies $\lim \limits_{x\to\infty} \frac{\epsilon(x)}{\operatorname{li}(x)} = 0$. My question is as follows:

Is anything known about an upper bound for $|p_n - \operatorname{li}^{-1}(n)|$? Here $p_n$ is the $n$-th prime.

This question was previously asked at math.SE, but this question received no clear answers.

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  • $\begingroup$ I was reading online about asymptotically $n$-th prime number by JUAN ARIAS DE REYNA, and i found this in the paper : if $\pi(x) = li(x)+O(\epsilon(x))$ then $p_x = li^{-1}(x) + O(\epsilon(x \ln(x)) \ln(x))$. I think this is what i am looking for, but i don't understand the $O$ notation, could any one give me precise answer. $\endgroup$
    – Ahmad
    Commented Mar 22, 2017 at 21:04
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    $\begingroup$ In analytic number theory, $O(f(x))$ denotes any function whose absolute value is less than some constant times $|f(x)|$. So if you have a good approximation $\pi(x)\approx\mathrm{li}(x)$, then you have a good approximation $p_n\approx\mathrm{li}^{-1}(n)$. It is straightforward to work out the relation of the two $O$-constants in the result you quote. $\endgroup$
    – GH from MO
    Commented Mar 22, 2017 at 21:52
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    $\begingroup$ So you can read off a concrete bound from my response to your earlier question: mathoverflow.net/questions/260959/… $\endgroup$
    – GH from MO
    Commented Mar 22, 2017 at 21:53
  • $\begingroup$ @GHfromMO yes, but i wanted to generalize the question to any given bound. $\endgroup$
    – Ahmad
    Commented Mar 22, 2017 at 22:25

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