Proving $\lim_{n \to \infty} \frac {\log{p_n}} {\log n} = 1$

Question

How do I show:

$$\lim_{n \to \infty} \frac {\log{p_n}} {\log n} = 1$$

where $p_n$ is the $n$th prime number without using the Prime Number Theorem?

Some context: The reason I can not use the PNT (or at least the form one might try to use) is because this is actually what I am trying to prove, or rather a certain form of the prime number theorem. The PNT states that $\pi(n) \sim \frac n {\log n}$, i.e.

$$\lim_{n \to \infty} \frac {\pi(n) \log n } n = 1$$

where $\pi(n)$ is the prime counting function.

Substituting $n \to p_n$ one has:

$$\lim_{n \to \infty}\frac {n \log p_n} {p_n} = 1$$

Now I would like to show that $p_n \sim n \log n$, i.e.

$$\lim_{n \to \infty} \frac {n \log n } {p_n} = 1$$

which requires the proof I am asking for.

Could you tell us your background. Along with where you encountered this question? — Gautam Shenoy, Dec 13 '15 at 14:41
Second thought... that could only be used in order to prove limit $\leq2$. — barak manos, Dec 13 '15 at 15:00
@Asvin A proof that this is implied by PNT would be most helpful. — user85798, Dec 13 '15 at 17:50
This page is quite good about general information on prime number estimates and provides references: https://primes.utm.edu/howmany.html — Asvin, Dec 13 '15 at 17:53
The wikipedia page is pretty good on information too - a little light on sources for minor details. — Asvin, Dec 13 '15 at 17:54

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

3

It suffices to prove the PNT up to a multiplicative constant. This is much easier than proving the PNT and was in fact done by Chebyshev.

Actually something slightly weaker suffices. A good enough upper bound can be extracted from the proof of Bertrand's postulate (not Bertrand's postulate, really the proof), as explained here; you get $p_n \le C n \log n$ for some constant $C$. And of course $p_n \ge n$ suffices for the upper bound.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 13 '15 at 18:35

Qiaochu Yuan

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Could you elaborate a bit more how do you get the bound for $p_n$ from the lemmas on the other reply you link? Because yo get ir I'm using also Bertrand's postulate and I don't think that is the intended way on your answer. – Eparoh Sep 27 '21 at 16:00

Proving $\lim_{n \to \infty} \frac {\log{p_n}} {\log n} = 1$

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