Bertrand’s conjecture, proof of

This is a version of Erdős’s proof as it appears in Hardy and Wright.

We begin with the following lemma.

Lemma.

Let $n$ be a positive integer and $p$ be a prime. The highest power of $p$ dividing $n\mathrm{!}$ is ${\displaystyle \mathrm{\sum }_j}\mathrm{\lfloor }{\displaystyle \frac{n}{p^j}}\mathrm{\rfloor }$, where $\mathrm{\lfloor }x\mathrm{\rfloor }$ is the floor of $x$.

Proof.

Let $p^k$ divide $n!$ with $k$ as large as possible. Every $p$th term of the sequence $1,\mathrm{\dots },n$ is divisible by $p$, so contributes a factor to $p^k$. There are $\lfloor {\displaystyle \frac{n}{p}}\rfloor$ such factors. Every $p^2$th term contributes an extra factor above that, providing $\lfloor {\displaystyle \frac{n}{p^2}}\rfloor$ new factors. In general, the $p^j$th terms contribute $\lfloor {\displaystyle \frac{n}{p^j}}\rfloor$ extra factors to $p^k$. So the highest power of $p$ dividing $n!$ is ${\displaystyle \sum _j}\lfloor {\displaystyle \frac{n}{p^j}}\rfloor$.∎

We now prove the theorem.

Bertrand’s conjecture.

Let $n>2$ be a minimal counterexample to the claim. Thus there is no prime $p$ such that .

In the sequence of primes

$$2,3,5,7,13,23,43,83,163,317,631,1259,2503,$$

each succeeding term is smaller than the double of its predecessor. This showsthat $n\ge 2503$.

The binomial expansion (http://planetmath.org/BinomialTheorem) of ${(1+1)}^{2n}$ has $2n+1$ terms and has largest term $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$. Hence

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)\ge \frac{4^n}{2n+1}\ge \frac{4^n}{{(2n)}^2}.$$

For a prime $p$ define $r(p,n)$ to be the highest power of $p$ dividing $\left(\genfrac{}{}{0pt}{}{2n}{n}\right)$. To compute $r(p,n)$, we apply the lemma to $(2n)!$ and $n!$. We get that

$$r(p,n)=\sum _{j\ge 1}\lfloor \frac{2n}{p^j}\rfloor -2\sum _{j\ge 1}\lfloor \frac{n}{p^j}\rfloor =\sum _{j\ge 1}\left(\lfloor \frac{2n}{p^j}\rfloor -2\lfloor \frac{n}{p^j}\rfloor \right).$$

The terms of the sum are all 0 or 1 and vanish when$j>\lfloor {\displaystyle \frac{\log (2n)}{\log p}}\rfloor$, so$r(p,n)\le \lfloor {\displaystyle \frac{\log (2n)}{\log p}}\rfloor$, that is,$p^{r(p,n)}\le 2n$.

Now $\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{n}}\right)={\displaystyle \prod _p}{p}^{r(p,n)}$. By the inequality just proved, primes larger than $2n$ do not contribute to this product, and by assumption there are no primes between $n$ and $2n$. So

$$\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\prod _{\begin{array}{c}1\le p\le n\\ p\text{ prime}\end{array}}{p}^{r(p,n)}.$$

For $n>p>\frac{2n}{3}$, $\frac{3}{2}>\frac{n}{p}>1$ and so for $p>\frac{2n}{3}>\sqrt{2n}$ we can apply the previous formula for $r(p,n)$ and find that it is zero. So for all $n>4$, the contribution of the primes larger than $\frac{2n}{3}$ is zero.

If $p>\sqrt{2n}$, all the terms for higher powers of $p$ vanish and $r(p,n)=\lfloor {\displaystyle \frac{2n}{p}}\rfloor -2\lfloor {\displaystyle \frac{n}{p}}\rfloor$.Since $r(p,n)$ is at most 1, an upper bound for thecontribution for the primes between $\sqrt{2n}$ and $\frac{2n}{3}$ is the product of all primes smaller than $\frac{2n}{3}$. This product is$\exp \left(\vartheta (\frac{2n}{3})\right)$, where $\vartheta (n)$ is the Chebyshev function

$$\vartheta (n)=\sum _{\begin{array}{c}p\le n\\ p\text{ prime}\end{array}}\log p.$$

There are at most $\sqrt{2n}$ primes smaller than $\sqrt{2n}$ and by the inequality $p^{r(p,n)}\le 2n$ their product is less than${(2n)}^{\sqrt{2n}}$. Combining this information, we get the inequality

$$\frac{4^n}{{(2n)}^2}\le \left(\genfrac{}{}{0pt}{}{2n}{n}\right)\le {(2n)}^{\sqrt{2n}}\exp \left(\vartheta (\frac{2n}{3})\right).$$

Taking logarithms and applying the upper bound of $n\log 4$ for $\vartheta (n)$ (http://planetmath.org/UpperBoundOnVarthetan), we obtain the inequality $\frac{n}{3}\log 4\le (\sqrt{2n}+2)\log (2n)$, which is false for sufficiently large $n$, say $n=2^{11}$. This shows that .

Since the conditions $n\ge 2503$ and are incompatible, there are no counterexamples to the claim.∎