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.

  • $\begingroup$ Could you tell us your background. Along with where you encountered this question? $\endgroup$ – Gautam Shenoy Dec 13 '15 at 14:41
  • $\begingroup$ @GautamShenoy Updated the question $\endgroup$ – user85798 Dec 13 '15 at 14:46
  • $\begingroup$ I think you can use the fact that $n<p_n<n^2$. $\endgroup$ – barak manos Dec 13 '15 at 14:57
  • $\begingroup$ Second thought... that could only be used in order to prove limit $\leq2$. $\endgroup$ – barak manos Dec 13 '15 at 15:00
  • $\begingroup$ I am guessing this is difficult then... $\endgroup$ – user85798 Dec 13 '15 at 17:33

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.


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