Enhancements of Bertrand's postulate?

Question

Is there a smallest real number $a$ such that there exist a natural number $N$ so that:

$n>N\implies p_{n+1}\leq a\cdot p_n$?

I believe it can be proved that $n>7\implies p_{n+1}\leq \sqrt 2\cdot p_n$.

Answer 1

There is no smallest such $a$, but the infimum of the set of such $a$ is $1$. In other words, for every $\epsilon>0$, there is a prime between $n$ and $(1+\epsilon)n$ for all sufficiently large $n$. This follows from the prime number theorem.

As a concrete example, for all $n\ge25$, there is a prime between $n$ and $\frac 65 n$ (Nagura 1952)..

Answer 2

Using the prime-number theorem: $$\begin{array} {rll} & p_{n+1} &\le a \cdot p_n \\ & { p_{n+1}\over p_{n}} &\le a \\ \text{(asymptotically:)}&{ (n+1) \log(n+1)\over n \log(n)} &\le a \\ & { (1+1/n) \over 1 } { \log(n+1)\over \log(n)} &\le a \\ & ( 1+1/n)\cdot (1+{ \log(1+1/n)\over \log(n)}) &\le a \end{array}$$ and the lhs on the last expression can be made arbitrarily near to $1$ by increasing $n$ so also $a$ can be taken arbitrarily small by using appropriate $n$

(However, that arguing/wording might be formally imprecise, I'm not much literate with formally proving; Hagen von Eitzen's arguing might be more correct)