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.

  • $\begingroup$ 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. $\endgroup$ Apr 7, 2013 at 3:49
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    $\begingroup$ 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 $\psi(2x)$? $\endgroup$
    – daniel
    Apr 7, 2013 at 5:44
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    $\begingroup$ 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. $\endgroup$ Apr 7, 2013 at 5:57
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    $\begingroup$ +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. $\endgroup$
    – daniel
    Apr 7, 2013 at 18:22


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