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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...

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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.

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    $\begingroup$ Is that right? $p_n$ converges to $\ n\ln(n)\ $ very slowly then... According to google, The 1 millionth prime number is $\ 15,485,863\ $ whereas $\ 1,000,000\ln(1,000,000)\ = 13,815,510$. That's not much closer (relatively speaking) than the $1000$-th prime number, $\ 7,919\ $ is to $1000\ln(1000)= 6908.\ $ So that's very slow convergence. But I guess it's not illegal for convergence to be very slow... $\endgroup$ May 21 at 23:39
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    $\begingroup$ Also, what is this elementary proof that the statement of the PNT that I gave is equivalent to $\frac{p_n}{n\ln{n}} \rightarrow 1$ ? [I believe you- but I still want to see the proof please]. The rest of what you say is easy to follow. $\endgroup$ May 21 at 23:47
  • $\begingroup$ It’s bad form to say that “$p_n$ converges to $n\log n,$” for $$p_n\sim n\log n.$$ If $x_n=n\log n$ then $$\frac{x_n}{\log x_n} =\dfrac{n}{1+\frac{\log\log n}{\log n}}$$ @AdamRubinson $\endgroup$ May 22 at 0:23
  • $\begingroup$ When $n=1000000,$ $\frac{\log\log n}{\log n}\approx 0.19.$ But clearly the fraction converges to $0.$ @AdamRubinson $\endgroup$ May 22 at 0:28
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    $\begingroup$ Well we have to be more precise by what we mean by "nearby" too. The asymptotics of $n$ primes by the first roughly $n \log n$ integers, does guarantee that for most $n$, the difference between the $n$-th prime and the $n+1$-th prime is $O(\log n)$. It however does not guarantee against the difference between the $n$th prime and the $n+1$-th prime being say $\sqrt{n}$ for an infinite number of $n$. $\endgroup$
    – Mike
    May 22 at 17:10
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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 $\lim_{n\to\infty}\frac{p_{n+1}}{p_n} = 1$.

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