Elementary proofs of prime gap theorems?

Question

"Obviously" it is thrue that $p_{n+1}<2p_n$. Testing for $n<10$ shows it is true for small $n$ and no mathematician or wannabe has ever doubt that it is true for big $n$. But there is no real simple arithmetic proof, so far, not using the prime number theorem or other results that isn't simple to prove.

So I wonder, are there any (non trivial) prime gap theorems at all with simple proofs? Or prime recursion inequalities of the form $p_{n+1}<f(p_1,p_2,\dots,p_n)$?

Can you prove: $p_{n+1}<p_n^2$ without using PNT, Bertrand's postulate,..?

Can you find a refinement of $\displaystyle p_{n+1}\le\prod^n_{i=1}p_i+1$, about just as simple to prove?

Answer 1

For the question about $p_{k+1} < p_k^2$, this should be fairly easy to deduce from the estimate $\sum_{\text{$p$ prime, }p \leq n} \frac{\ln p}{p} = \ln n + O(1)$, so long as you're careful to make the $O(1)$ bound explicit. The proof of the estimate is elementary and short. (See the exercises to Chapter 2 of Vinogradov's Elements of Number Theory.)

Namely, if the $O(1)$ term is between $A$ and $B$, then as long as $n > e^{B-A}$, there must be a prime between $n$ and $n^2$. Then you can check small values of $n$ manually.

I haven't worked out the details of what $A$ and $B$ are, but I don't think they're large.

Edit: Here is a quick proof of Mertens' estimate, which is what we used above. On Wikipedia it is stated that $O(1)$ is bounded by $2$ in absolute value, so this would establish the theorem for all $n > e^4$.

This proof can be immediately generalized to prove that if $n$ is large enough, then there is necessarily a prime number between $n$ and $n^{1 + \varepsilon}$.

Answer 2

There is a refinement of $\displaystyle p_{n+1}\le\prod^n_{i=1}p_i+1$, just about as simple to prove:

$ \displaystyle p_n\leq 2\prod_{q\in\Bbb P'}^{q<p_n}q+1$, where $\Bbb P'=\{p\in\Bbb P|\exists m\in\Bbb N:p=4m+3\}$and $n>3$.

Proof: Let $ \displaystyle A_n=2\prod_{q\in\Bbb P'}^{q<p_n}q+1$ and suppose $p_n>A_n$. Then $A_n$ only has prime factors $p_k<p_n$ such that $p_k=4m+1,\; m\in\Bbb N$. Thus $A_n=4m'+1,\; m\in\Bbb N$ which is obviously not true.
Hence $p_n\leq A_n$.

A simple proof for the corresponding conjecture with $\Bbb P''=\{p\in\Bbb P|\exists m\in\Bbb N:p=4m+1\}$ is probably also possible and it might be possible to show for any similar class of primes for $n$ big enough.