# Easy proof of falsehood of $\pi(n) \leq C \cdot \text{ln}(n)$ for the prime counting function $\pi$

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?

## 1 Answer

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.

• 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 – American Igor Apr 17 at 10:09
• Another way is $-\log (s-1) \sim \log \zeta(s) \sim \sum_p p^{-s} = s \int_1^\infty \pi(x) x^{-s-1}dx$ – reuns Apr 17 at 10:18
• @reuns: Can you elaborate? – American Igor Apr 17 at 10:23
• $\pi(x) = O(\ln x)$ means the integral converges for $\Re(s) > 0$ contradicting its singularity at $s=1$ – reuns Apr 17 at 10:36