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I am a bit stuck on how to derive the following estimate for the $n$-th prime.

$$p_n = n \log n + n \log \log n + \mathcal{O} (n)$$

(We are given that $\pi (x) = Li(x) + \mathcal{O}(\exp(-a \sqrt{\log x}))$ for some constant $a$.)

I am a bit stuck, and looking online, I just seem to find references to a very old paper by I.M. Pervushin (which I cannot actually find), and this result seems to just be assumed in a lot of literature.

Can we derive this directly from just the PNT: $\pi(x) \sim \frac{x}{\log x}$?

Any help or hints would be appreciated.

Alibabaexpress
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    See this duplicate, with references to Rosser and Dusart. See also the answers here. – Dietrich Burde Dec 07 '19 at 15:04
  • @DietrichBurde This is very clearly not a duplicate; your first reference is a question which simply asks for formulae for the nth prime, and the answer just notes that the formula in my question is a consequence of PNT. The second is all about the estimate $p_n \sim n \log n$ which is different to the estimate in my question. – Alibabaexpress Dec 07 '19 at 15:27
  • See your title. Precise estimates for $p_n$ are given, in modern language, by Dusart in his thesis. This gives what you want. In the duplicate this is given: $n \ln n + n(\ln\ln n - 1) < p_n < n \ln n + n \ln \ln n$ for $n\ge{}6$. Hence $p_n = n \log n + n \log \log n + \mathcal{O} (n)$. – Dietrich Burde Dec 07 '19 at 15:28
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    @DietrichBurde Well the title is exactly that - it tells you roughly what the question is going to be about. A very quick read of my question makes it clear that I am asking a specific question on how to derive a specific estimate rather than a free for all on estimates for $p_n$. (And yes, proofs of stronger estimates are obviously technically still valid proofs, but they are not in the spirit of the question. Dusart’s thesis, for example, would not even constitute a proof of my question since it assumes the statement of my question which had been proved about 100 years prior.) – Alibabaexpress Dec 07 '19 at 15:33
  • Then Rosser's and Schönfeld's proof is the right one for you. It is $(2.20)$ here. The proof is direct. – Dietrich Burde Dec 07 '19 at 15:34
  • @DietrichBurde In the “duplicate”, you are correct that the answer states a fact which implies the statement in my question, but the whole point of the question is that I’m looking for a way to prove it directly. – Alibabaexpress Dec 07 '19 at 15:34
  • Yes, so $(2.20)$ in Rosser and Schoenfeld, page $66$. – Dietrich Burde Dec 07 '19 at 15:36
  • @DietrichBurde Thanks for the reference, but given that this reference is not even mentioned in either of the “duplicates” and given that the actual questions in the duplicates are clearly distinct from my question (though on a similar topic), I would hardly call my question a duplicate. – Alibabaexpress Dec 07 '19 at 15:39
  • I haven't voted for duplicate, don't worry. However, the article by Rosser in fact is mentioned there. – Dietrich Burde Dec 07 '19 at 15:40
  • @DietrichBurde I am looking through the article you linked, and though it states the fact that I am trying to prove, it doesn’t explicitly prove it (I may have to go down a rabbit hole of further reference chasing). Anyhow, I have managed to work out a proof for myself by now... I may post an answer in a bit. – Alibabaexpress Dec 07 '19 at 15:43

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