Was curious about this question, can't seem to find this on the internet, perhaps my googling skills are rustly lol
Let $(P_k)_{k\geq1}$ be the sequence of prime numbers where the $k$-th term represents the $k$-th prime. So we have $(P_k)_{k\geq1}=2,3,5,7,11,13,\dots$ and so on.
Now, consider the limits
\begin{align*}
\lim_{n\to\infty}\dfrac{P_n}{P_{n+1}}\text{ and }\lim_{n\to\infty}\dfrac{P_{n+1}}{P_{n}}.
\end{align*}
Does either one of these limit exists? I know as number gets larger, the gap between primes gets larger in general but is the gap close enough between consecutive prime numbers in a way that these limits approaches, say $1$?
1, If we can show there exists, say our favorite greek symbol $\delta$, such that between any $n\in\mathbb{N}$, we can always find some prime number $p_i\in(n-\delta,n+\delta)$, then this limit obviously approaches 1. But I don't think we can assume such thing.
2, Recall some theorem showed that between any $n$ and $2n$ (perhaps with some kind of limitation) there is a prime between them, so that means if the limit exists, then $\lim_{n\to\infty}\dfrac{P_n}{P_{n+1}}$ is bounded below by $1/2$ and the other one bounded above by $2$.
Best Answer
It should exist and should be well-known. Here is a sloppy calculation. Recall that by PNT, $\pi(n)=\frac{n}{\ln n}(1+o(1))$. Thus, $n^{th}$ prime $p_n$ is roughly of order $n\ln n$. It then follows that, $\frac{p_{n+1}}{p_n}= \frac{p_{n+1}}{(n+1)\ln (n+1)}\frac{(n+1)\ln (n+1)}{n\ln n}\frac{n\ln n}{p_n}$ has the limit one, since $\frac{p_{n+1}}{(n+1)\ln(n+1)}=1+o(1)$ and $\frac{n\ln n}{p_n}=1+o(1)$.