I can easily and shortly prove that given $ε$, computable using all primes less than $N$, there will be at least one prime number between $n$ and $(1+ε)n$, where $n > N$.
It proves Bertrand's Postulate for $ε = 1$. And it gives a better result for $ε < 1$.
So my questions are:
- Is this result notable?
- Should I publish it?
- If so, where and how?
Thanks for your answers/comments.
It is hard to evaluate a statement that makes extensive use of the principle of elimination of quantifiers - in the sense of "I don't write the quantifiers, and you deal with it" -; however, in the information you've provided there is a passage, right in the beginning, that stands out:
Ah. So the meticulous reader should understand that $\varepsilon$ is not actually given (as in a statement which begins with "$\forall \varepsilon$"), but rather dependent on $N$, as in $$\forall N,\exists \varepsilon(N),\cdots$$
Such a statement alone does not allow to consider the case $\varepsilon=1$, because as far as you know $\varepsilon(N)$ may never be smaller than $2000$.
In principle, one may argue whether the statement proceeds with "$\forall n>N$" or "$\exists n>N$"; however, since the latter gives a trivial theorem, let's assume that we're facing: $$\forall N,\exists\varepsilon(N),\forall n>N,\exists p(\varepsilon(N),n)\text{ prime},\ n<p(\varepsilon(N),n)<(1+\varepsilon(N))n$$
This looks like the theorems linked by user Patrick Stevens in the comments, but it's actually not the same thing: those results should rather be phrased as (refinements of) $$\forall \varepsilon>0,\exists N(\varepsilon),\forall n>N(\varepsilon),\exists p(\varepsilon,n)\text{ prime},\ n<p(\varepsilon,n)<(1+\varepsilon)n$$ which is stronger than your claim. Specifically, it can be recovered from yours only by further showing that there is some choice of $\varepsilon$ such that $\lim_{N\to\infty}\varepsilon(N)=0$: if that's the case, you can select $N(\varepsilon)=\min\{ M\in\Bbb N\,:\, \varepsilon(M)<\varepsilon\}$.