One can show via the PNT that $$\lim_{n\to \infty} \frac{\pi((1+\epsilon)n) - \pi(n)}{n/\log{n}} = \epsilon,$$
for any $\epsilon > 0$, which in particular implies that for any $\epsilon > 0$ and for any $n$ big enough there is always a prime between $n$ and $(1+\epsilon)n$.
Bertrand's postulate is a special case of this $(\epsilon = 1)$ with the added benefit that it holds for all $n > 1$, so it works for all positive integers (excluding 1).
What's the smallest known $\epsilon > 0$ such that there is always a prime between $n$ and $(1+\epsilon)n$? Clearly we must have $\epsilon > 1/2$ so that $(1+\epsilon)2 > 3$, but i'm not sure $1/2$ works.
If you want it to work for all $n\geq 2$ then let $\epsilon_n = \frac{p_{n+1}}{p_n}-1$ and take the maximum $n$, from PNT we know that $ \lim \limits_{n \to \infty}\frac{p_{n+1}}{p_n}=1 $ and so $ \lim \limits_{n\to \infty} \epsilon_n =0$ and checking for smaller cases we find that $\epsilon_2 = \frac{5}{3}-1 \approx 0.666$ is the minimum epsilon that the interval $ [n ,n(1+\epsilon)]$ is garneted to contains a prime