Help understanding limits in Ramanujans proof of Bertrand's Postulate

Question

I cannot see where $3\sqrt{x}$ comes from in the limits

$$\begin{equation}\vartheta(x)-\vartheta(\frac{1}{2}x)>\frac{1}{6}x-3\sqrt{x}\text{, if }x>300.\end{equation}$$

I understand where:

$\log[x]! - 2\log[\frac{1}{2}x]! < \frac{3}{4}x$ if $x > 0$

$\psi(x)-\psi(x/2)< \frac{3}{4}x$ if $x > 0$

$\log[x]! - 2\log[\frac{1}{2}x]! > \frac{2}{3}x$ if $x > 300$

$\psi(x)-\psi(x/2)+\psi(x/3)> \frac{2}{3}x$ if $x > 300$

$\psi(x)\geq{x/2} x$ for $x > 150$

comes from, but given

$$\psi(x)-\psi(x/2)+\psi(x/3)\leq\nu(x)+2\psi(\sqrt{x})-\nu(x/2)+\psi(x/3)<\nu(x)-\nu(x/2)+x/2+3\sqrt{x}$$

I cannot see where the $3\sqrt{x}$ from in his 1919 paper 'A proof of Bertrand’s postulate' http://www.zyymat.com/ramanujans-proof-of-bertrands-postulate.html.

Answer 1

The key is the inequality in line (13) to which he refers in line (14).

$\psi(x)< \frac{3x}{2}$ if $x > 0. $

Focusing on the right hand side of your last inequalities.

$$\nu(x)+2\psi(x^{1/2})-\nu(x/2)+\psi(x/3)<\nu(x)-\nu(x/2)+x/2+3\sqrt{x}, $$

note that we can subtract $\nu(x)-\nu(x/2)$ from both sides leaving

$$2\psi(x^{1/2})+\psi(x/3)<x/2+3\sqrt{x}. $$

But by (13),

$$2\psi(x^{1/2})< 2\cdot3\sqrt{x}/2 = 3\sqrt{x}, $$

and $$\psi(x/3) <3(x/3)/2 = x/2. $$

In short, it's a matter of using (13) and changing the functions accordingly.