Lemma. \[\prod_{p\le n} p \le 4^{n-1}.\]
Remark. In fact, the real answer is closer to \(e^{1.000028}\) or something. See wikipedia page for Chebyshev prime function (which is the \(\ln\) of this function).
Proof. proof of lemma is by induction. We use the almost-central binomial coefficient: \[\binom{2k+1}{k}\le 4^{k},\] because note that \[\prod_{p\in (k,2k+1]} p \le \binom{2k+1}{k}.\] That is, primes \(p\in (k,2k+1]\) show up exactly once in \(\binom{2k+1}{k}\) because they can’t be cancelled by the denominator. Wow that’s neat. Anyways, then you do the induction and it works.
Theorem. There are at least \((\frac{2}{3}-o(1)) \frac{n}{\log n}\) primes between \([n,2n]\) for large enough \(n\).
Proof. Now we use central binomial coefficient for our bound so cool. key observation 1: \[\nu_p(\binom{2n}{n}) \le \log_p n\] I have another blog post about this fact. Key observation 2: for \(p\in (\frac{2n}{3}, n)\) \[\nu_p(\binom{2n}{n}) = 0.\] Key observation 3: For \(p> \sqrt{2n}\) \[\nu_p(\binom{2n}{n}) \le 1.\]
Let \(f(n)\) be the number of primes \(p\in [n,2n]\). We have the bound
\[\frac{4^{n}}{n} \le \binom{2n}{n} \le \prod_{p\le \sqrt{2n}}2n \prod_{p\in (\sqrt{2n}, 2n / 3)} p \prod_{p\in [n,2n]}p\]
Using the lemma from the beginning of this blog post we get
\[\frac{4^{n}}{n} \le (2n)^{\sqrt{2n}} 4^{\frac{2n}{3}} (2n)^{f(n)}.\] Taking logs and stuff gives the desired bound.