In 1949, Leo Moser gave a proof of  Bertrand’s postulate in the paper “A theorem of the distribution of primes”(American Mathematical Monthly,624-624,1949,vol56). Maybe this proof  is the simplest one among all proofs of  Bertrand’s postulate.

1. If $$n<p\leqslant 2n$$, then $$p$$ occurs exactly once in $${2n\choose n}$$;
2. If $$n\geqslant 3, \frac{2n}3<p<n$$, then $$p$$ doesn’t occurs in $${2n\choose n}$$;
3. If $$p^2>2n$$, then $$p$$ occurs at most once in $${2n\choose n}$$;
4. If $$2^\alpha\leqslant 2n<2^{\alpha+1}$$, then $$p$$ occurs at most $$\alpha$$ times in $${2n\choose n}$$.

If  $$2^\alpha<2n\leqslant 2^{\alpha+1}$$, suppose there is no prime $$p$$ with

$$n<p<2n,$$

then

$${2n\choose n}\leqslant{2a_1\choose a_1}{2a_2\choose a_2}\dotsm{2\choose 1}\bigg({2a_k\choose a_k}{2a_{k+1}\choose a_{k+1}}\dotsm{2\choose 1}\bigg)^\alpha,$$

where $$a_1=\big[\frac{n+1}3\big], k=\big[\frac{\alpha+1}2\big]$$, and $$a_i=\big[\frac{a_{i-1}+1}3\big]$$, for $$i=2,\,3,\,\dotsc\,.$$

By $$1,\,2,\,3$$ and $$(1)$$, every prime which appears on the left-hand side of $$(2)$$ appears also on the right; and those primes which appear with multiplicity greater than $$1$$ on the left appear on the right with multiplicity at least $$2\alpha+1$$, which by $$4$$ is at least equal to the multiplicity with which they appear on the left.

On the other hand, It is obvious that

$$2n>2a_1+2a_2+\dotsb+ 2+\alpha(2a_k+2a_{k+1}+\dotsb+2)$$

for $$n>2^{11}.$$ Hence the inequality in $$(2)$$ should be reversed. So for $$n>2^{11}$$, we have a contradiction which proves Bertrand’s postulate for these values of $$n$$.

Sylvester’s theorem  the binomial coefficient

$${n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$

has a prime divisor $$p$$ greater than $$k$$. In other words, if $$n\geq 2k$$, then the product of  $$k$$ consecutive integers greater than $$k$$ is divisible by a prime greater than $$k$$. Note that for $$n=2k$$ we obtain precisely Bertrand’s postulate.

Theorem

1. if $$n>6$$ is an integer, then there always exists at least two prime numbers $$p$$, one  is the form $$4k+1$$ and the other $$4k+3$$, such that $$n<p<2n$$.
2. if $$k$$ is a positive  integer, then there always exists a positive  integer $$N$$ such that  for every  integer $$n>N$$,  there exist at least $$k$$ prime numbers $$p$$ such that $$n<p<2n$$.
• 对任意正整数 $$n>6$$ ,  至少存在一个 $$4k+1$$ 型和一个 $$4k+3$$  型素数 $$p$$  使得  $$n<p<2n$$.
• 对任意正整数 $$k$$,  存在正整数 $$N$$,  使得只要正整数 $$n>N$$,  就存在 $$k$$ 个素数 $$p$$ 使得 $$n<p<2n$$.

Bertrand’s postulate states that if $$x\geq4$$, then there always exists at least one prime $$p$$ with$$x<p<2x-2$$. A weaker but more elegant formulation is: for every $$x>1$$ there is always at least one prime p such that $$x<p<2x$$.

In 1919, Ramanujan used properties of  the  Gamma function to give a simple  proof  that  appeared as  a  paper “A proof of Bertrand’s postulate” in the  Journal of the Indian Mathematical Society $$11: 181–182$$.

Let’s begin with the  Chebyshev function :

The  first  Chebyshev  function $$\vartheta(x)$$  is defined with

$$\vartheta(x)=\sum_{p\leqslant x}\log p,$$

The second Chebyshev function $$\psi(x)$$ is given by

$$\psi(x)=\sum_{i\geqslant 1}\vartheta(x^{\frac1i}),$$

so

$$\log \left[x\right]!=\sum_{i\geqslant 1}\psi(\frac1ix) ,$$

where $$\left[x\right]$$denotes as usual the greatest integer $$\leqslant x$$.

From $$(2)$$ we have

$$\psi(x)-2\psi(\sqrt{x})=\vartheta(x)-\vartheta(x^\frac12)+\vartheta(x^\frac13)-\dotsb,$$

and from $$(3)$$

$$\log\left[x\right]!-2\log\left[\frac12x\right]!=\psi(x)-\psi(\frac12x)+\psi(\frac13x)-\dotsb .$$

Remembering  that $$\vartheta(x)$$ and $$\psi(x)$$ are steadily increasing functions, from $$(4)$$ and $$(5)$$ we get that

$$\psi(x)-2\psi(\sqrt{x})\leqslant\vartheta(x)\leqslant\psi(x);$$

and

$$\psi(x)-\psi(\frac12x)\leqslant \log \left[x\right]!-2\log \left[\frac12x\right]!\leqslant\psi(x)-\psi(\frac12x)+\psi(\frac13x).$$

But it is easy to see that

$$\begin{split}\log \Gamma(x)-2\log \Gamma(\frac12x+\frac12)&\leqslant \log \left[x\right]!-2\log \left[\frac12x\right]!\\&\leqslant\log \Gamma(x+1)-2\log \Gamma(\frac12x+\frac12).\end{split}$$

Now using Stirling’s approximation we deduce from $$(8)$$ that

$$\log \left[x\right]!-2\log \left[\frac12x\right]!<\frac34x, \text{if} x>0;$$

and

$$\log \left[x\right]!-2\log \left[\frac12x\right]!>\frac23x, \text{if} x>300.$$

It follows from $$(7)$$, $$(9)$$ and  $$(10)$$ that

$$\psi(x)-\psi(\frac12x)<\frac34x, \text{if} x>0;$$

and

$$\psi(x)-\psi(\frac12x)+\psi(\frac13x)>\frac23x, \text{if } x>300.$$

Now changing $$x$$ to $$\frac12x$$, $$\frac14x$$, $$\frac18x$$, $$\dotsc$$ in $$(11)$$ and adding up all the results, we obtain

$$\psi(x)<\frac32x,\text{if} x>0.$$

Again we have

$$\begin{split}\psi(x)-\psi(\frac12x)+\psi(\frac13x)&\leqslant\vartheta(x)+2\psi(\sqrt{x})-\vartheta(\frac12x)+\psi(\frac13x)\\&<\vartheta(x)-\vartheta(\frac12x)+\frac12x+3\sqrt{x},\end {split}$$

in virtue of  $$(6)$$ and $$(13)$$.

It  follow from $$(12)$$ and $$(14)$$ that

$$\vartheta(x)-\vartheta(\frac{1}{2}x)>\frac{1}{6}x-3\sqrt{x}\text{, if }x>300.$$

But it is obvious that

$$\frac16x-3\sqrt{x}\geqslant 0\text{, if } x\geqslant 324,$$

Hence

$$\vartheta(2x)-\vartheta(x)>0, \text{if } x\geqslant162.$$

In other words there is at least one prime between $$x$$ and $$2x$$ if  $$x\geq162$$. Thus Bertrand’s postulate is proved for all values of  $$x$$ not less than $$162$$; and, by actual verification, we find that it is true for smaller values.

It is obvious that

$$\vartheta(x)-\vartheta(\frac12x)\leqslant(\pi(x)-\pi(\frac12x))\log x,$$

for all values of  $$x$$. It follows from $$(15)$$ and $$(18)$$ that

$$\pi(x)-\pi(\frac12x)>\frac1{\log x}(\frac16x-3\sqrt{x})\text{, if} x>300.$$

From this we easily deduce that

$$\pi(x)-\pi(\frac12x)\geqslant1,2,3,4,5,\dotsc\text{, if } x\geqslant2,11,17,29,41,\dotsc$$

respectively.