Improved Betrand's postulate

Question

I want to show that $2p_{n-2} \geq p_{n}-1$...

Bertand's postulate shows us that $4p_{n-2}\geq p_{n}$ but can we improve on this?

any ideas?

Answer 1

There are a lot of improvements of the Bertrand postulate but yes, in this case it's sufficient to use that there is a prime between $n$ and $6n/5$, $n $ sufficiently large (as you write $n \geq 25$). So we have $$p_{n-2}<p_{n-1}<\frac{6p_{n-2}}{5} $$ and $$ p_{n-1}<p_{n}<\frac{6p_{n-1}}{5}=p_{n-1}+\frac{p_{n-1}}{5} $$ then $$p_{n}<\frac{36p_{n-2}}{25}\Rightarrow p_{n}-1<\frac{25p_{n}}{18}<2p_{n-2}. $$