3

The thread here lists some pretty unique, simple, and non-trivial proofs on lower bounds for the prime counting function. We can also ask the analogous question:

What is the simplest to prove, non-trivial upper bound for the prime counting function ?

I consider trivial to be something along the lines of $\pi(n) \le n/2$ or worse bounds.

Community
  • 1
Sandeep Silwal
  • 8,135
  • It's easy to show that we eventually have $\pi(n) \le Cn + D$ for any $C$ and some $D$ depending on $C$. The actual upper bound, around $\frac{n}{\log n}$, is not much smaller than this... – Qiaochu Yuan Aug 18 '16 at 19:13
  • @QiaochuYuan: I agree. However, I was hoping to collect some cute proofs like in the thread I linked. – Sandeep Silwal Aug 18 '16 at 19:18

1 Answers1

4

Chebyshev's theorem is a weak version of the PNT, and the upper bound $$ \forall x\geq x_0,\qquad \pi(x)\leq (2\log 2)\cdot\frac{x}{\log x} \tag{1}$$ is not so difficult to prove with a subtle argument. From Stirling's approximation/Wallis product we have that the central binomial coefficient behaves in the following way: $$ C_n=\binom{2n}{n}\sim\frac{4^n}{\sqrt{\pi n}}\tag{2} $$ but every prime in the range $(n,2n)$ is a divisor of $C_n$, hence $$ \sum_{n<p<2n}\log p \leq \log C_n \leq n\log 4\tag{3} $$ and by considering the intervals $(n,2n)$, $\left(\frac{n}{2},n\right)$,$\left(\frac{n}{4},\frac{n}{2}\right)$ and so on we get: $$ \vartheta(x) = \sum_{p<x}\log p \leq x\log 4.\tag{4} $$ Now it is enough to apply summation by parts to get $(1)$.

Jack D'Aurizio
  • 353,855