I saw a proof that $|p_n - li^{-1}(n)| \leq n e^{-c \sqrt{\ln(n)}} $, without saying anything about $c$ !
My questions is, what the explicit value of $c$ ??
It just says for some number $c$ without giving any information about $c$ !
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$\begingroup$ What's $p_n$ and $li^{-1} (n)$? $\endgroup$– Pat DevlinCommented Jan 30, 2017 at 23:15
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$\begingroup$ Presumably the $n$th prime and the inverse function of $li$, the logarithmic integral: $li(x)=\int_2^x 1/\log x\,dx$. $\endgroup$– Anthony QuasCommented Jan 30, 2017 at 23:20
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5$\begingroup$ Where did you see a proof? $\endgroup$– StoppleCommented Jan 30, 2017 at 23:23
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2$\begingroup$ Don't close this question as I gave a nice answer to it :-) $\endgroup$– GH from MOCommented Jan 31, 2017 at 5:58
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1 Answer
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An explicit version of the Prime Number Theorem required here is Theorem 1.12 on Page 42 of Dusart's thesis. In particular, it follows from this theorem that $$ |\pi(x)-\mathrm{Li}(x)|<x e^{-0.32\sqrt{\ln x}},\qquad x\geq 59. $$ Using Kadiri's explicit zero-free region, Trudgian sharpened this result to $$ |\pi(x)-\mathrm{Li}(x)|<x e^{-0.39\sqrt{\ln x}},\qquad x\geq 229.$$
Strictly speaking, these inequalities do not answer the original question. On the other hand, applying either of them at $x=p_n$, one obtains a similar bound for $|n-\mathrm{Li}(p_n)|$ in terms of $n$ (using a crude but concrete Chebyshev type bound $p_n\asymp n\log n$), and then applying $\mathrm{Li}^{-1}$ to both $n$ and $\mathrm{Li}(p_n)$, one concludes an analogous bound for $|\mathrm{Li}^{-1}(n)-p_n|$.
Added. See arXiv:2206.12557 and the references therein for even sharper results.
answered Jan 31, 2017 at 3:38