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
\begin{equation}\vartheta(x)=\sum_{p\leqslant x}\log p,\end{equation}
The second Chebyshev function \(\psi(x)\) is given by
\begin{equation}\psi(x)=\sum_{i\geqslant 1}\vartheta(x^{\frac1i}),\end{equation}
so
\begin{equation}\log \left[x\right]!=\sum_{i\geqslant 1}\psi(\frac1ix) ,\end{equation}
where \(\left[x\right]\)denotes as usual the greatest integer \(\leqslant x\).
From \((2)\) we have
\begin{equation}\psi(x)-2\psi(\sqrt{x})=\vartheta(x)-\vartheta(x^\frac12)+\vartheta(x^\frac13)-\dotsb,\end{equation}
and from \((3)\)
\begin{equation}\log\left[x\right]!-2\log\left[\frac12x\right]!=\psi(x)-\psi(\frac12x)+\psi(\frac13x)-\dotsb .\end{equation}
Remembering that \(\vartheta(x)\) and \(\psi(x)\) are steadily increasing functions, from \((4)\) and \((5)\) we get that
\begin{equation}\psi(x)-2\psi(\sqrt{x})\leqslant\vartheta(x)\leqslant\psi(x);\end{equation}
and
\begin{equation}\psi(x)-\psi(\frac12x)\leqslant \log \left[x\right]!-2\log \left[\frac12x\right]!\leqslant\psi(x)-\psi(\frac12x)+\psi(\frac13x). \end{equation}
But it is easy to see that
\begin{equation}\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}\end{equation}
Now using Stirling’s approximation we deduce from \((8)\) that
\begin{equation}\log \left[x\right]!-2\log \left[\frac12x\right]!<\frac34x, \text{if} x>0;\end{equation}
and
\begin{equation}\log \left[x\right]!-2\log \left[\frac12x\right]!>\frac23x, \text{if} x>300.\end{equation}
It follows from \((7)\), \((9)\) and \((10)\) that
\begin{equation}\psi(x)-\psi(\frac12x)<\frac34x, \text{if} x>0;\end{equation}
and
\begin{equation}\psi(x)-\psi(\frac12x)+\psi(\frac13x)>\frac23x, \text{if } x>300. \end{equation}
Now changing \(x\) to \(\frac12x\), \(\frac14x\), \(\frac18x\), \(\dotsc\) in \((11)\) and adding up all the results, we obtain
\begin{equation}\psi(x)<\frac32x,\text{if} x>0.\end{equation}
Again we have
\begin{equation}\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}\end{equation}
in virtue of \((6)\) and \((13)\).
It follow from \((12)\) and \((14)\) that
\begin{equation}\vartheta(x)-\vartheta(\frac{1}{2}x)>\frac{1}{6}x-3\sqrt{x}, \text{ if }x>300.\end{equation}
But it is obvious that
\begin{equation}\frac16x-3\sqrt{x}\geqslant 0\text{, if } x\geqslant 324,\end{equation}
Hence
\begin{equation}\vartheta(2x)-\vartheta(x)>0, \text{if } x\geqslant162.\end{equation}
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.
Since \(\pi(x)-\pi(\frac12x)\) is the number of primes bwteen \(x\) and \(\frac12x\), and \(\vartheta(x)-\vartheta(\frac12x)\) is the sum of logarithms of primes bwteen \(x\) and \(\frac12x\), It is obvious that
\begin{equation}\vartheta(x)-\vartheta(\frac12x)\leqslant(\pi(x)-\pi(\frac12x))\log x,\end{equation}
for all values of \(x\). It follows from \((15)\) and \((18)\) that
\begin{equation}\pi(x)-\pi(\frac12x)\gt\frac1{\log x}(\frac16x-3\sqrt{x}), \text{ if } x\gt 300.\end{equation}
From this we easily deduce that
\begin{equation}\pi(x)-\pi(\frac12x)\geqslant1,2,3,4,5,\dotsc\text{, if } x\geqslant2,11,17,29,41,\dotsc\end{equation}
respectively.