# Is there a smallest real number a such that there exist a natural number N so that: n>N->p_(n+1)<=a x p_n?

Is there a smallest real number $a$ such that there exist a natural number $N$ so that:
$n>N\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{p}_{n+1}\le a\cdot {p}_{n}$?
I believe it can be proved that $n>7\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}{p}_{n+1}\le \sqrt{2}\cdot {p}_{n}$.
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Jazmyn Bean
There is no smallest such $a$, but the infimum of the set of such $a$ is 1. In other words, for every $ϵ>0$, there is a prime between $n$ and $\left(1+ϵ\right)n$ for all sufficiently large $n$. This follows from the prime number theorem.
As a concrete example, for all $n\ge 25$, there is a prime between $n$ and $\frac{6}{5}n$
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Gorlandint
Using the prime-number theorem:
$\begin{array}{rll}& {p}_{n+1}& \le a\cdot {p}_{n}\\ & \frac{{p}_{n+1}}{{p}_{n}}& \le a\\ \text{(asymptotically:)}& \frac{\left(n+1\right)\mathrm{log}\left(n+1\right)}{n\mathrm{log}\left(n\right)}& \le a\\ & \frac{\left(1+1/n\right)}{1}\frac{\mathrm{log}\left(n+1\right)}{\mathrm{log}\left(n\right)}& \le a\\ & \left(1+1/n\right)\cdot \left(1+\frac{\mathrm{log}\left(1+1/n\right)}{\mathrm{log}\left(n\right)}\right)& \le a\end{array}$
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 $n$