Strengthening Bertrand's postulate using the prime number theorem

Question

In its page on Bertrand's postulate Wikipedia says

It follows from the prime number theorem that for any real $\varepsilon >0$ there is an $n_0 > 0$ such that for all $n > n_0$ there is a prime $p $ such that $n < p < ( 1 + ε )n$ .

I need to quote this result for $\varepsilon = 1/2$. I'd like a reference more formal than Wikipedia.

Answer 1

The ariticle on wikipededia cites that with a foot note (as of the time I write this) #8 refering to: G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, p. 494.

from Wikipedia

It follows from the prime number theorem that for any real $ {\displaystyle \varepsilon >0}$ there is a ${\displaystyle n_{0}>0} $ such that for all ${\displaystyle n>n_{0}} $ there is a prime $ {\displaystyle p}$ such ${\displaystyle n<p<(1+\varepsilon )n}$. It can be shown, for instance, that

${\displaystyle \lim _{n\to \infty }{\frac {\pi ((1+\varepsilon )n)-\pi > (n)}{n/\log n}}=\varepsilon ,} $[8]

[8]G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008, p. 494.

Answer 2

tersely worded, but Dusart gives, for $x \geq 396738,$ a prime between $x$ and $$ x \left( 1 + \frac{1}{25 \log^2 x} \right) $$

I think it was on the arxiv, let me find it.

Yes, this is Proposition 6.8.

For your fixed target $3x/2,$ you may fill in the smaller numbers with the table of maximal prime gaps at https://en.wikipedia.org/wiki/Prime_gap#Numerical_results . I went through the thing once by machine, for $p \geq 11$ and $p < 4 \cdot 10^{18},$ we get gap $g < \log^2 p.$