Prove $p$$_n$$_+$$_1$ $<$ $2p_n$ without using the Bertrand's Postulate

Question

Recently I have been researching on the Bertrand's Postulate to find and elementary proof of it. I have been able to prove that (if I have not made a very pathetic mistake) for any composite $n$ $(\geq5)$, there exists at least one prime between $n$ and $2n$ (and that proof isn't similar to any of the earlier 'standard' proofs). Not only that, in each case I have found the prime which should be within the bound. But I am a bit skeptic about my proof, so I will post it later. However, where I am stuck at is regarding the proof of $p$$_n$$_+$$_1$ $<$ $2p_n$. Any suggestion as to prove this inequality (using elementary methods) without making use of the Bertrand's Postulate?

Answer 1

The inequality is equivalent to the Bertrand Postulate, so it is pretty hard to prove it without it.

Here is why:

BP implies the inequality:

Indeed, by BP, there exists a a prime number between $p_n$ and $2p_n$. This shows that

$$p_{n+1} < 2 p_n $$

The inequality implies BP

Let $n$ be any positive number, and let $p_{k}$ be the largest prime which is less or equal than $n$. Then, by the inequality $p_{k+1}<2 p_k$, we have

$$p_{k} \leq n < p_{k+1} < 2p_{k} \leq 2n$$

Therefore there exists a prime ($p_{k+1}$) strictly between $n$ and $2n$.