About primes in an interval

Question

I find this result without a proof:

The prime number theorem implies that for any $\epsilon > 0$, there exists an integer $N$ such that for all $x > N$, there exists a prime between $x$ and $x(1 + \epsilon)$.

I am asking about a simple proof for this result.

What have you tried? What other proofs might be relevant? Does the infinitude of primes guarantee this result immediately? — David G. Stork, Oct 16 '19 at 17:44
@DavidG.Stork: In fact I am tried the fact that $N(x)∼(x/(log x)), x→∞$ but without any sucess. — Safwane, Oct 16 '19 at 17:46
Using the prime number theorem, how many primes do you expect between $x$ and $x(1+\epsilon)$ in terms of $x$ and $\epsilon$? How does this value behave as $x\to\infty$? — 79037662, Oct 16 '19 at 17:54
I think, Dusarts bounds for the number of primes below $x$ should be sufficient for a proof. — Peter, Oct 16 '19 at 18:09
No, the Dusart bounds are proven for sufficiently large $x$ without relying on some unproven conjecture. — Peter, Oct 16 '19 at 18:12
I only read this in wikipedia, but I did not find a reference of a proof of Dusart's bounds. Anyway, you probably look for something easier. — Peter, Oct 16 '19 at 18:15

score 2 · Accepted Answer · answered Oct 16 '19 at 18:17

By prime number theorem, there are approximately $\frac{N}{\ln(N)}$ primes less than $N$, for sufficiently large $N$. Therefore, between $x$ and $x(1+\epsilon)$ there are approximately $$\frac{x(1+\epsilon)}{\ln(x(1+\epsilon))}-\frac{x}{\ln(x)}$$ primes. This expression is equivalent to $$\frac{x(1+\epsilon)\ln(x)-x\ln(x(1+\epsilon))}{\ln(x(1+\epsilon))\ln(x)}$$ which, by L'Hopital's rule, approaches $\infty$ as $x\to\infty$.

In particular, this means there is some $N$ such that for all $x>N$ we have at least one prime between $x$ and $x(1+\epsilon)$.

About primes in an interval

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