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What is the best known estimatio for $\limsup |\frac{p_{n+1}}{n+1}-\frac{p_n}{n}|$ ?

I know about prime gaps, Prime Number Theorem etc. famous results, but this is something I don't know how to search info about. Does this problem have any name?

user454221
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    $\lim \sup |\frac{p_{n+1}}{n+1}- \frac{p_n}{n}| = 0$ using PNT, stronger results makes it go faster to $0$. –  Nov 17 '20 at 22:52
  • Is it really true? $p_n\sim n\ln n$ and $p_{n+1}\sim(n+1)\ln(n+1)$, so $\frac{p_{n+1}}{n+1}\sim\ln(n+1)$ and $\frac{p_{n}}{n}\sim\ln(n)$, but can we simply conclude that the difference is $\sim\ln(n+1)-\ln n$ ? If yes, please explain how, I dont see it. – user454221 Nov 17 '20 at 23:12
  • This approach is messy, you need to bound $p_{n+1}$ using $p_n$. –  Nov 17 '20 at 23:15
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    @Ahmad you will always get a pesky $\varepsilon$ you will have to carry. – rtybase Nov 17 '20 at 23:17
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    @user454221 it is definitely $<1$, using this or this. – rtybase Nov 17 '20 at 23:18

1 Answers1

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A result from PNT is that there is prime number in the interval $ [x , x+ \frac{x}{25\ln^2 x}]$ for all $x \geq 396738$.

So we have that $ p_n < p_{n+1} < p_n + \frac{p_n}{25\ln^2 p_n} \leq p_n + \frac{n}{\ln n}$

Since $ n (\ln n+ \ln \ln n-1) < p_n < n(\ln n+\ln \ln n)$ for all $n \geq 6$

So $|\frac{p_{n+1}}{n+1}-\frac{p_n}{n}| \leq \frac{p_{n+1}-p_n}{n+1} +p_n |\frac{1}{n+1}-\frac{1}{n}| \leq \frac{\frac{n}{\ln n}}{n}+ \frac{p_n}{n^2} \leq \frac{1}{\ln n}+ \frac{2 n \ln n}{n^2} = \frac{1}{\ln n}+\frac{2\ln n}{n} \to 0$ as $n \to \infty$

Note : all the inequalities and lemmas derived from PNT are found in Dusart's paper

rtybase
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  • Did you assume $p_{n+1}-p_n\leq \frac{n}{\ln{n}}$? – rtybase Nov 17 '20 at 23:26
  • Its not an assumption, its corollary from PNT @rtybase , the first line gives stronger upper bound –  Nov 17 '20 at 23:27
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    Have a look at this. I'd love to see a proof. – rtybase Nov 17 '20 at 23:31
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    @rtybase a proof of what?,, also OP question is very related to $\frac{g_n}{n}$ not to $g_n$ itself and since Baker result gives that $ g_n << n^{0.6}$ so $\frac{g_n}{n} \to 0$ as $n\to \infty$. –  Nov 17 '20 at 23:36
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    A proof of $p_{n+1}-p_n\leq \frac{n}{\ln{n}}$ please ;) – rtybase Nov 17 '20 at 23:39
  • I know $\frac{g_n}{n}\to 0, n\to\infty$ btw. Stated in the comments above. – rtybase Nov 17 '20 at 23:41
  • @rtybase from PNT we know that $| x- \theta(x)| \leq \frac{x}{\ln^3 x}$ for instance, actually way better bound are out there, and we know that $ \theta(p_{n+1}) - \theta(p_n) > 0$ , you just need to convince yourself that $ \theta(p_n +\frac{n}{\ln n}) -\theta(p_n)>0$ –  Nov 17 '20 at 23:44
  • @rtybase the above is just a sketch for a proof i saw many times in a lot of papers, also the best bound using PNT is $ |x -\theta(x)| \leq x e^{-c (\ln x)^{0.6}}$ for small number $c$ –  Nov 17 '20 at 23:47
  • @rtybase note that it is enough to prove $p_{n+1}-p_n\leq \frac{2n}{\ln{n}}$ and that is fairly trivial since $p_{n+1}-p_n \le p_n$ etc – Conrad Nov 17 '20 at 23:47
  • Yes, sooo ... how long it takes to add these details or links to the details? As I wrote, I could have stopped at $\frac{g_n}{n}\to 0$ with the link to the relevant result. Or should I edit the answer? – rtybase Nov 17 '20 at 23:52