Let $\pi(n)$ be the number of primes in the range $1,\dotsc,n$.

The following statement is true: There is no $C>0$ such that $\pi(n) \leq C \cdot \text{ln}(n)$ for all $n\geq 1$.

It follows immediately from the prime number theorem which is a much stronger result.

Still, since the above statement is much weaker than the PNT, I was wondering if it has a simple proof.

Is there a proof of the above theorem which is simpler than the known proofs of the prime number theorem?


There are all kinds of rougher estimates, for example you can use Chebyshev's estimate to show that $\pi (x) > c x/ \log (x) $ for a positive c and large enough x.

  • $\begingroup$ Thank you. I looked for the proof. It is indeed much easier than PNT's. Nevertheless, it is still much stronger than what I need, so maybe there are even simpler proofs of what I need $\endgroup$ – American Igor Apr 17 at 10:09
  • $\begingroup$ Another way is $-\log (s-1) \sim \log \zeta(s) \sim \sum_p p^{-s} = s \int_1^\infty \pi(x) x^{-s-1}dx$ $\endgroup$ – reuns Apr 17 at 10:18
  • $\begingroup$ @reuns: Can you elaborate? $\endgroup$ – American Igor Apr 17 at 10:23
  • $\begingroup$ $\pi(x) = O(\ln x)$ means the integral converges for $\Re(s) > 0$ contradicting its singularity at $s=1$ $\endgroup$ – reuns Apr 17 at 10:36

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