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.

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