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.