A result on prime numbers [closed]

Hi everyone

This is my first post... I do mathematics from home... ie., not attached with any institution... I have deduced some results...

$\lim \inf_{n\to\infty} \frac{d_n}{\log p_n} = 0$

and, for constants $A,B$

$\lim_{n\to\infty} \log p_n - \sum_{i=1}^{n-1} \frac{d_i}{p_{i+1}} = A$

$\lim_{n\to\infty} \log p_n - \sum_{i=1}^{n-1} \frac{d_i}{p_{i}} = B$

Where, $p_n$ is the nth prime... $d_n = p_{n+1} - p_n$

My question is: Do you think these results are good ?

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closed as not a real question by Anton GeraschenkoJan 1 '10 at 21:12

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Your result is due to Goldston and Yildirim, and was only established in the last 10 years, using sophisticated sieve theory techniques. It would be very impressive if you had found an easy/elementary proof of this. –  Thomas Bloom Jan 1 '10 at 15:26
I am no specialist in analytic number theory, but from what I know from analysis the limit computation (10) is clearly wrong. You have the product of two factors, the first (p_n/log p_n) of which goes to infinity and the second goes to zero, and so it does not follow that the product goes to zero. –  Leonid Positselski Jan 1 '10 at 17:20
@Leonid, I think you're right and you should post your observation as an answer -- it could be accepted then. –  Ilya Nikokoshev Jan 1 '10 at 18:17
@rpg: your first implication is not justified. This is called "1 to the power \infty", a kind of expression whose limit you cannot know just from knowing the limits of the constitutient subexpressions. You are repeating your original mistake here, in a slightly disguised form. –  Leonid Positselski Jan 1 '10 at 19:11
It's fine to ask, "are these results known and does somebody have a reference?" But this has turned into a game of "debug my proof for me," which (in this form at least) is not appropriate for Math Overflow. I'm closing the question (direct discussion about it to tea.mathoverflow.net/discussion/124). If the question is edited to sharpen it (i.e. make it a question that has an answer), or if somebody otherwise convinces me (or another moderator, or five 3k+ rep users), the question can be reopened. –  Anton Geraschenko Jan 1 '10 at 21:12

Disclaimer: I am no specialist in Analytic Number Theory, nor did I read the whole paper under the link. I just looked into the end of the argument, and there is a limit computation (10) there.

From what I know from Analysis, this computation is clearly wrong, not in the sense that the answer is necessarily wrong, but in the sense that the premises do not justify the conclusion. The author attempts to compute the lower limit of the product $$\liminf_{n\to\infty}\left(\frac{p_n}{\log p_n}\log\frac{p_{n+1}}{p_n}\right)$$ as the product of the limits. He replaces the second factor with $log(1)=0$ and proceeds to claim that the lower limit of the product is $0$. However, even though the (lower) limit of the second factor may well be $0$, the limit of the first factor is clearly $\infty$, so one cannot compute the lower limit of the product in this way.

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check the updated paper :) arxiv.org/pdf/0912.4646 Cheers!!! –  Roupam Ghosh Jan 1 '10 at 19:03
Leonid above correctly says that you are now computing $1^\infty$: you can't just claim it's equal to 1. @Leonid, I think it's more reasonable to move discussion here. @rpg16, once you arrive to understanding with Leonid, please accept his answer! –  Ilya Nikokoshev Jan 1 '10 at 19:17
@Ilya: unfortunately I perceive very little point in continuing the discussion. Let me simply state that my answer applies to the prepint arxiv:0912.4646 version 4 and not to any later or other version. –  Leonid Positselski Jan 1 '10 at 19:32
@rpg16: Looking at the new answer, I don't think it is correct. You do correctly move the $p_n$ term inside the log, but now this ratio to the $p_n$ can end up being quite large (I don't think the limit is 1, in fact I think--but am no expert on this subject--that there are results on what this limit is). So your next step replacing it all by a 1 is probably incorrect. –  Ben Weiss Jan 1 '10 at 19:43
OMG, I did not realize that I was commenting on a version of a paper that was not yet permanently saved. The author replaced it, and now there are no traces left of what I have commented on in my answer. I should have known better than falling prey to that. –  Leonid Positselski Jan 1 '10 at 19:56

I would expect the first limit to diverge like $\log\log n$ ... Here is why: we know that $p_n \sim n \log n$ while on average $d_i \sim \log i$ so i would expect your second limit to be about $\log n + \log\log n - \sum_{i=1}^n{(\log i)/(i \log i)} \sim \log\log n$ ... On the other hand if you can prove that this limit is bounded I think that would be a surprising result by itself! I'd be happy to hear what others have to say about it...

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