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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?

Brad Graham
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  • I have to go now, but I'm wondering if we could show that $2p_{n-2} <p_{n}$ - is false for $n=3$, and then disprove the rest by induction? (Not sure if that kind of proof is plausible) – Brad Graham Aug 04 '15 at 20:54
  • I just read that Nagura (1952) proved that for all $n$ bigger then 25, there exists a prime between $n$ and $1.2 \times n$. Is this true? If so, then $p_{n} \leq 1.44p_{n-2}$ for all $n$ bigger then 25, so showing that the improved betranda postulate holds for primes less than 25, is sufficient to prove it right, right? – Brad Graham Aug 04 '15 at 22:55
  • See the Wikipedia article Bertrand's Postulate , especially the section Better Results. – DanielWainfleet Aug 05 '15 at 02:28

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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 $$p_{n}<\frac{36p_{n-2}}{25}\Rightarrow p_{n}-1<\frac{25p_{n}}{18}<2p_{n-2}. $$

Marco Cantarini
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