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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?

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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.

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  • $\begingroup$ 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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