There are various proofs of Bertrand's postulate. There is quite an easy one available as a follow-up to an easy version if one treats it together with the proof of the usual (double) Chebyshev bound , which I give below in tex. You need to know that \vartheta(x) = \sum_{p \leq x}\log(p)\psi(x) = \sum_{p^k \leq x}\log(p)and that T(x) = \sum_{n \leq x}\log(n)Here is the Chebyshev bound...
\begin{Chebyshev} The inequalities \log(2)x - \log(4x) \leq \psi(x) \leq 2\log(2)x + 2\log(x) hold for $x \geq 1$.\end{Chebyshev}
\begin{proof} The terms in the last sum in the computation \begin{align} T(x) - 2T\left(\frac{x}{2}\right) &=
\sum_{n \leq x}\log(n) - 2\sum_{m \leq x/2}\log(m) \ &= \sum_{n \leq x}\log(n) - 2\sum_{2m \leq x}\log(2m) + 2\sum_{2m \leq x}\log(2) \ &= \sum_{n \leq x}(-1)^{n-1}\log(n) + 2\left[\frac{x}{2}\right]\log(2) \end{align} alternate in sign and increase in magnitude. So \left|T(x) - 2\,T\left(\frac{x}{2}\right) - 2\left[\frac{x}{2}\right]\log(2)\right| \leq \log([x]) for $x \geq 1$. Thus \log(2)x - \log(4x) \leq T(x) - 2\,T\left(\frac{x}{2}\right) \leq
\log(2)x + \log(x). Substituting the expression T(x) = \sum_{n \leq x}\psi\left(\frac{x}{n}\right) into $T(x) - 2\,T(x/2)$ yields \psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) - \cdots = T(x) - 2\,T\left(\frac{x}{2}\right). Then \psi(x) \geq \log(2)x - \log(4x) since $\psi$ is an increasing and nonnegative function. And \psi(x) - \psi\left(\frac{x}{2}\right) \leq \log(2)x + \log(x) for the same reason. Adding up all the inequalities \psi\left(\frac{x}{2^j}\right) - \psi\left(\frac{x}{2^{j+1}}\right) \leq \log(2)2^{-j}x + \log\left(\frac{x}{2^j}\right) for $j = 0,1,2,\ldots,[\log(x)/\log(2)]$ gives \psi(x) \leq 2\log(2)x + \left[\frac{\log(x)}{\log(2)}\right]\log(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} since $\psi(x/2^{j+1}) = 0$ when $x/2^{j+1} < 1$. Now assume that the inequality \psi(x) \leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) holds for as a positive integer $k$. It certainly holds for $k = 1$unit. Apply it on One optimizes the right hand side of \psi(x) \leq \psi\left(\frac{x}{2}\right) + \log(2)x + \log(x) to obtain \begin{align} \psi(x) &\leq 2\log(2)\frac{x}{2} + \frac{\log^2(x/2)}{\log(2)} - (k-1)\log\left(\frac{x}{2}\right) + \log(2)x + \log(x) \ &= 2\log(2)x + \frac{\log^2(x)}{\log(2)} - k\log\left(\frac{x}{2}\right). \end{align} Then \begin{align} \psi(x) &\leq 2\log(2)x + \frac{\log^2(x)}{\log(2)} - \left[\frac{\log(x)}{\log(2)}\right] \log\left(\frac{x}{2}\right) \ &\leq 2\log(2)x + 2\log(x) \end{align} holds by repeated halving.\end{proof}
The advantages proof of this version are two: It does not require Stirling's formula, and it is just right the Chebyshev bound for subsequently proving Bertrand's Postulate by the method of Ramanujan . Here goes...
\begin{Bertrand} For every $x \geq 2$ there exists at least one prime $p$ with $x/2 < p \leq x$.\end{Bertrand}
\begin{proof} The basic idea is to use the difference $\vartheta(x) - \vartheta(x/2)$ to detect primes in the interval $(x/2,x]$. A suitable positive lower bound for $\vartheta(x) - \vartheta(x/2)$ is going to imply that there is at least one prime in $(x/2,x]$ for all $x$ sufficiently large. At the end, the intervals $(x/2,x]$ for small values of $x$ have to be treated separately.
The definitions of $\psi$ and $\vartheta$ yield \psi(x) &= \sum_{p^k \leq x}\log(p) = \sum_{k=1}^{\infty}\sum_{p^k \leq x}\log(p) \\ &= \sum_{k=1}^{\infty}\sum_{p \leq x^{1/k}}\log(p) = \vartheta(x) + \vartheta(x^{1/2}) + \vartheta(x^{1/3}) + \cdotsand so \psi(x) - 2\psi(x^{1/2}) = \vartheta(x) - \vartheta(x^{1/2}) + \vartheta(x^{1/3}) - \cdots.The inequality $\psi(x) - 2\psi(x^{1/2}) \leq \vartheta(x) \leq \psi(x)$ follows since $\vartheta$ is an increasing and nonnegative function.Fetch the inequality \psi(x) - \psi\left(\frac{x}{2}\right) + \psi\left(\frac{x}{3}\right) \geq T(x) - from the proof of the Chebyshev bound. Retention (getting rid of the term $\psi(x/3)$ is an idea due to Ramanujan. Now \vartheta(x) - \vartheta\left(\frac{x}{2}\right) &\geq \psi(x) - 2\psi(x^{1/2}) - \psi\left(\frac{x}{2}\right) \\ &\geq T(x) - 2T\left(\frac{x}{2}\right) - \psi\left(\frac{x}{3}\right) - 2\psi(x^{1/2}) \\ &\geq \log(2)x - \log(4x) - 2\log(2)\frac{x}{3} \\ &- 2\log\left(\frac{x}{3}\right) - 4\log(2)x^{1/2} - 4\log(x^{1/2})by the Chebyshev bound and its proof. So $\vartheta(x) - \vartheta(x/2) \geq f(x)$ with f(x) = \log(2)x/3 - 4\log(2)x^{1/2} - 5\log(x) + The derivative f'(x) = \log(2)/3 - 2\log(2)x^{-1/2} - 5x^{-1}is visibly an increasing function. Then $f'(x) > 0$ on the interval $[347,\infty)$ since $f'(347) = 0.14$. Thus $f(x)$ is increasing on this interval and hence positive there because $f(347) = 0.09$. So there is a prime in every interval $(x/2,x]$ with $x \geq 347$. Now a trick due appeal to E. G. H. Landau completes the proof: Each term in Stirling's formula at the chain $2,3,5,7,11,17,29,53,97,179,347$ of primes is smaller than twice its predecessor. So there is also a prime in each interval $(x/2,x]$ for $2 \leq x \leq 347$same time).
\end{proof}
