Is there a simple way to prove Bertrand's postulate from the prime number theorem?

Solution 1:

I'm not sure if it's what you're looking for, but the proof of Bertrand's postulate is somewhat related to the proof of Chebyshev's theorem, which says that $\pi(x) = \Theta(x/\log x)$; both proofs rely on careful estimation of the central binomial coefficients $\binom{2n}{n}$.

To prove Chebyshev's theorem, one can start by finding the estimate

$$ \frac{2^{2n}}{2n+1} < \binom{2n}{n} < 2^{2n}, \tag{1} $$

then using it to conclude that $\vartheta(x)/x < 4\log 2$ and hence

$$ \limsup_{x \to \infty} \frac{\pi(x)}{x/\log x} = \limsup_{x \to \infty} \frac{\vartheta(x)}{x} \leq 4 \log 2, $$

and $\psi(x)/x > (x-2)\log 2/x - \log(x+1)/x$ and hence

$$ \liminf_{x \to \infty} \frac{\pi(x)}{x/\log x} = \liminf_{x \to \infty} \frac{\psi(x)}{x} \geq \log 2, $$

where $\vartheta$ and $\psi$ are Chebyshev's functions.

To prove Bertrand's postulate, one can improve the estimate in $(1)$ to

$$ \frac{2^{2n}}{2\sqrt{n}} < \binom{2n}{n} < \frac{2^{2n}}{\sqrt{2 n}}, \tag{2} $$

which can be used to show that

$$ \vartheta(2n) - \vartheta(n) > \left(\frac{2n}{3}-1\right)\log 2-\left(\frac{\sqrt{2n}+1}{2}\right)\log 2 > 0$$

for $n \geq 2^6$. This is exactly Bertrand's postulate for $n \geq 2^6$. The remaining cases are checked by hand.

My reference for these proofs is Chandrasekharan's Introduction to Analytic Number Theory.