Does the prime number theorem tell us that the next prime number is guaranteed to be relatively nearby?

Question

Let $\ p_n\ $ be the $\ n$-th prime number.

Does the prime number theorem ,

$\Large{\lim_{x\to\infty}\frac{\pi(x)}{\left[ \frac{x}{\log(x)}\right]} = 1},$

imply that:

$ \displaystyle\lim_{n\to\infty}\ \frac{p_n}{p_{n+1}} = 1\ ?$

Edit: I totally get where the vote-to-closes come from and I kind of agree with them. Yeah this is not the question I intended to ask actually. I think I've done an X-Y communication thingy. I'll leave the question and accept the answer though. But I have learned something about prime numbers along the way in reading the answers...

Answer 1

Indeed, yes! It can be shown elementarily that the statement of the PNT that you gave is equivalent to $\frac{p_n}{n\ln{n}} \rightarrow 1$. Since $\frac{n+1}{n} \rightarrow 1$, $\frac{\ln(n+1)}{\ln{n}}\rightarrow 1$, it follows that $\frac{p_{n+1}}{p_n} \rightarrow 1$.

Edit: here’s the elementary proof. $\frac{\pi(p_n)\ln{p_n}}{p_n} \rightarrow 1$, thus $p_n+o(p_n)=(n\ln{p_n})$ (so $n=o(p_n)$). Write $q_n=\frac{p_n}{n}$. Then $p_n+o(p_n)=(n\ln{n})+n\ln{q_n}$. Now, $\frac{q_n}{\ln{p_n}} \rightarrow 1$, and $p_n \rightarrow \infty$, so that $\ln{q_n}=\ln{\ln{p_n}}+o(1)$. Thus $n\ln{q_n}=o(p_n)$, so that $p_n+o(p_n)=n\ln{n}$, QED.

Answer 2

You can also use results from the prime gap problem. Here, as $\lim_{n\to\infty} \frac{g_n}{p_n} = 0$ and $g_n = p_{n+1} - p_n$, you can conclude that $\lim_{n\to\infty}\frac{p_{n+1}}{p_n} = 1$.