most recent 30 from http://mathoverflow.net 2013-05-26T05:48:20Z http://mathoverflow.net/feeds/question/10377 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10377/a-result-on-prime-numbers Roupam Ghosh 2010-01-01T15:20:49Z 2010-01-01T19:50:27Z

<p>Hi everyone</p> <p>This is my first post... I do mathematics from home... ie., not attached with any institution... I have deduced some results...</p> <p><code>$\lim \inf_{n\to\infty} \frac{d_n}{\log p_n} = 0$</code></p> <p>and, for constants <code>$A,B$</code></p> <p><code>$\lim_{n\to\infty} \log p_n - \sum_{i=1}^{n-1} \frac{d_i}{p_{i+1}} = A$</code></p> <p><code>$\lim_{n\to\infty} \log p_n - \sum_{i=1}^{n-1} \frac{d_i}{p_{i}} = B$</code></p> <p>Where, <code>$p_n$</code> is the nth prime... <code>$d_n = p_{n+1} - p_n$</code></p> <p>My question is: Do you think these results are good ?</p>

http://mathoverflow.net/questions/10377/a-result-on-prime-numbers/10391#10391 Leonid Positselski 2010-01-01T18:59:14Z 2010-01-01T18:59:14Z

<p>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.</p> <p>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.</p>

http://mathoverflow.net/questions/10377/a-result-on-prime-numbers/10395#10395 maki 2010-01-01T19:50:27Z 2010-01-01T19:50:27Z

<p>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...</p>