Extrema of the Ratio of Consecutive Primes

Question

Let $p_i$ denote the $i$th prime number. We know that $\frac{p_{n+1}}{p_n}\rightarrow 1$ as $n\rightarrow\infty$. Therefore, if we pick some real number $c>1$, there should be some positive integer $N$ such that $\frac{p_{n+1}}{p_n}<c$ for all $n>N$. Do we know anything about these values of $N$ for different values of $c$? For example, if we let $c=\frac{3}{2}$, do we know the positive integer $N$ (or an estimate) such that $\frac{p_{n+1}}{p_n}<c$ for all $n>N$. We could ask a similar question for other values of $c$ such as $\sqrt{2}$ or $\frac{5}{3}$. Thank you.

Answer 1

This question (or rather, an equivalent one) is the subject of my paper with V. Shevelev and P. Moses. One of the more useful tools is the 2010 paper of Dusart which uses a large verification of zeta zeros to get numerical bounds on many prime-related functions. In particular he shows that:

• For $x\ge396738,$ the interval $[x,x+x/(25\log^2x)]$ contains at least one prime.
• The $n$-th prime is at least $n\left(\log n+\log\log n-1+\frac{\log\log n-2.1}{\log n}\right)$ for $n\ge3$.
• The $n$-th prime is at most $n\left(\log n+\log\log n-1+\frac{\log\log n-2}{\log n}\right)$ for $n\ge688383$.

So if $n\ge688383$ then the gap between $p_n$ and $p_{n+1}$ is at most $$ \frac{n\left(\log n+\log\log n-1+\frac{\log\log n-2}{\log n}\right)}{25\log^2\left(n\left(\log n+\log\log n-1+\frac{\log\log n-2}{\log n}\right)\right)} $$

and hence $p_{n+1}/p_n-1$ is at most $$ f(n)=\frac{\log n+\log\log n-1+\frac{\log\log n-2}{\log n}}{25\left(\log n+\log\log n-1+\frac{\log\log n-2.1}{\log n}\right)\log^2\left(n\left(\log n+\log\log n-1+\frac{\log\log n-2}{\log n}\right)\right)} $$

which you can evaluate numerically.

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As an example, $f(688383)\approx0.0001533<0.0001534$ and so if $n\ge688383$ then $p_{n+1}/p_n<1.0001534.$ Checking the primes below that point you can actually improve this to $n>49414$ which is sharp since $p_{49415}/p_{49414}=1+98/604073\approx1.0001622.$