# Understanding Ramanujan's approach in his proof of Bertrand's Postulate

I've been reading through Ramanujan's proof of Betrand's Postulate and I'm not clear why he didn't state his proof in terms of $\varphi(2x) - \varphi(x)$

What would be wrong with this approach for example:

Let $\varphi(x) = \sum_{p\le x}\log p$

Let $\psi(x) = \sum_{n \ge 1}\varphi(x^{\frac{1}{n}})$

$\psi(2x) - 2\psi(\sqrt{2x}) = \varphi(2x) - \varphi([2x]^\frac{1}{2}) + \varphi([2x]^{\frac{1}{3}}) - \ldots$

$\psi(2x) - 2\psi(\sqrt{2x}) \le \varphi(2x) \le \psi(2x)$

$\psi(2x) - \psi(x) \le \log[2x]! - 2\log[x]! \le \psi(2x) - \psi(x) + \psi(\frac{2}{3}x)$

$\psi(2x) - \psi(x) + \psi(\frac{2}{3}x) \le \varphi(2x) - \varphi(x) + 2\psi(\sqrt{2x}) + \psi(\frac{2}{3}x)$

Using $\psi(x) < \frac{3}{2}x$, if $x > 0$ from (13) in the proof, we have:

$\log[2x]! - 2\log[x]! \le \varphi(2x) - \varphi(x) + 3\sqrt{2x} + x$

Using Stirling's formula, we have $2x! > \sqrt{4\pi{x}}(\frac{2x}{e})^{2x}$ and $\exists{w}$ so that $x! < \sqrt{w\pi{x}}(\frac{x}{e})^x$

If I did my calculations correctly: $\log[2x]! - 2\log[x]! > \frac{4}{3}x$ for $x \ge 48$

For $x > 162$, $\frac{1}{3}x - 3\sqrt{2x} > 0$

So for $x \ge 162$, we have:

$\varphi(2x) - \varphi(x) > 0$

edit: I've modified the argument to be closer to Ramanujan's proof based on Daniel's comment. Before I used $\psi(2x) - \psi(x) \le \log[2x]! - \log[x]! \le \psi(2x) - \psi(x) + \psi(x)$ which may not be valid.

• Hi @Daniel, Ramanujan does his analysis using $\varphi(x) - \varphi(\frac{1}{2}x)$ and then uses the Gamma function to deal with the lower bound. My thought is that his argument requires less steps if he uses $\varphi(2x) - \varphi(x)$ from the beginning. As I understand it, he would then not need to use the Gamma function to establish the lower bound. He could use Stirling's formula for factorials. – Larry Freeman Apr 7 '13 at 3:49
• We have $\psi(2x)+\psi(2x/2)+\psi(2x/3)...-\psi(x)-\psi(x/2)-\psi(x/3)...$ So $\psi(2x)-\psi(x) < \psi(2x)$ and $\psi(2x/2)-\psi(x/2)< \psi(2x/2)...$ but the sum of these differences could exceed $\psi(2x)$? – daniel Apr 7 '13 at 5:44
• Great point. This may be the reason that Ramanujan took the approach that he did! Ramanujan's proof stems from $\log[x]! - 2\log[\frac{1}{2}x]! \le \psi(x) - \psi(\frac{1}{2}x) + \psi(\frac{1}{3}x)$. This may not hold for the approach that I used. Let me investigate more. – Larry Freeman Apr 7 '13 at 5:57
• +1. i didn't check the stirling approx but plotted the functions. if there's a mistake i don't see it. nice question either way. – daniel Apr 7 '13 at 18:22