Bertrand's postulate

Put $P(y):={\prod }_{p\le y}p$ for real $y\ge 2$. We claim that $P(y)\le 4^{y-1}(y\ge 2)\mathrm{.}$ Let $q$ be the largest prime with $q\le y$. Then $P(y)=P(q)$ and $4^{q-1}\le 4^{y-1}$, so it suffices to prove the claim when $y=q$ is prime. For $q=2$ this is immediate. Let $q=2m+1$ be an odd prime, and assume inductively that $P(r)\le 4^{r-1}$ for every integer $r$ with $2\le r\le 2m$. Then $P(q)=P(m+1){\prod }_{m+1<p\le 2m+1}p\mathrm{.}$ Every prime in the second product divides $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)=\frac{(2m+1)!}{m!(m+1)!}$ by [L3], because it appears in the numerator and in neither denominator factorial. Also [L3] gives ${\sum }_{k=0}^{2m+1}\left(\genfrac{}{}{0px}{}{2m+1}{k}\right)=2^{2m+1},$ and the two equal middle terms $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)=\left(\genfrac{}{}{0px}{}{2m+1}{m+1}\right)$ therefore satisfy $\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)\le 2^{2m}$. Thus $P(q)\le P(m+1)\left(\genfrac{}{}{0px}{}{2m+1}{m}\right)\le 4^m\cdot 2^{2m}=4^{2m}=4^{q-1}\mathrm{.}$ Taking logarithms and using [L4], we obtain $\theta (y)\le 2y\log 2(y\ge 2)\mathrm{.}$

Assume now that $n\ge 468$, and put $Q_n:={\prod }_{n<p\le 2n}p\mathrm{.}$ By [L2], every prime in this interval divides $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$, so $Q_n$ is a factor of $\left(\genfrac{}{}{0px}{}{2n}{n}\right)$. Also, if $2n\mathrm{/}3<p\le n$, then $\lfloor 2n\mathrm{/}p\rfloor =2$, $\lfloor n\mathrm{/}p\rfloor =1$, and $p^2>{\left(\frac{2n}{3}\right)}^2>2n$ because $n\ge 468>9\mathrm{/}2$. Hence [L2] gives $v_p\left(\genfrac{}{}{0px}{}{2n}{n}\right)=0$ for that range. Therefore $\log \left(\genfrac{}{}{0px}{}{2n}{n}\right)\le \log Q_n+\theta (2n\mathrm{/}3)+U_n,$ where $U_n:={\sum }_{k=2}^{\lfloor {\log }_2(2n)\rfloor }\theta ((2n{)}^{1\mathrm{/}k})\mathrm{.}$ Indeed, for $p\le 2n\mathrm{/}3$ each summand $\lfloor \frac{2n}{p^k}\rfloor -2\lfloor \frac{n}{p^k}\rfloor$ in [L2] is at most $1$, so the remaining logarithmic contribution is bounded by the $k=1$ layer $\theta (2n\mathrm{/}3)$ together with the higher prime-power layers $U_n$.

The remaining range $2\le n\le 467$ is finite. A direct scan on September 1, 2026 checked each interval $(n,2n)$ and found a prime witness in every case; for example the last few witnesses are $463<467<926,464<467<928,467<479<934.$ So the statement also holds throughout the residual finite range.

Step 1.1 implies $\theta (y)\le 2(\lfloor y\rfloor +1)\log 2$ for every real $y\ge 2$. Hence $U_n\le 2(\sqrt{2n}+1)\log 2+2\lfloor {\log }_2(2n)\rfloor ((2n{)}^{1\mathrm{/}3}+1)\log 2.$ For $n\ge 468$ one has $\lfloor {\log }_2(2n)\rfloor \le (2n{)}^{1\mathrm{/}3}$. Applying step 1.1 at $\lfloor 2n\mathrm{/}3\rfloor +1$, combining the resulting bounds with step 1.2 and the lower bound from [L1], and then simplifying gives $\log Q_n\ge \frac{2}{3}n\log 2-4\log 2-\log (2n+1)-2\log 2((2n{)}^{2\mathrm{/}3}+\sqrt{2n}+(2n{)}^{1\mathrm{/}3}+1)\mathrm{.}$ The right-hand side is positive at $n=468$ and has positive derivative for every $n\ge 468$, so $Q_n>1$ throughout that range. Hence some prime $p$ satisfies $n<p\le 2n$, and the endpoint $p=2n$ is impossible because $2n$ is even and larger than $2$. Thus $n<p<2n$ for every $n\ge 468$.

Steps 2.1 and 1.3 together prove Bertrand's postulate for every integer $n>1$.