Can anyone prove or disprove the following statement?
$ \lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{\sqrt{p_n}} = 0.$
Can anyone prove or disprove the following statement?
$ \lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{\sqrt{p_n}} = 0.$
At present, no one can prove or disprove this statement. It is very famous and still open. The best unconditional result is Baker-Harman-Pintz:
$\lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{p_n^{0.525}} < \infty.$
Quite likely it can be shown that this limit is exactly $0$, but I haven't read enough of the paper to be certain.
I don't think anyone knows, although I am looking up stuff just in case. Meanwhile, what people suspect is the Cramer-Granville conjecture, $ \lim \sup \frac{p_{n+1}-p_n}{\left( \log p_n \right)^2} = 2 e^{- \gamma} = 1.1229\ldots, $ where the logarithm is to base $e = 2.718281828459\ldots$ and $\gamma = 0.5772156649\ldots$ is the Euler-Mascheroni constant. This conjecture, and the Baker result mentioned in the other answer, are in GRANVILLE PDF and WookiePedia. Hmmm, not quite, Granville mentions the earlier $0.535$ result of Baker and Harman. With Pintz they later got it to $0.525.$
This is consistent with a (mostly) stronger conjecture that I made up for no good reason except that it also applies to small numbers, $ p_{n+1} \, - \, p_n < \; 3 \; \log^2 \, p_n. $ For example, $p_1 = 2,\; \log 2 = 0.693147\ldots, \log^2 \, 2 = (0.693147\ldots)^2 = 0.480453\ldots, \; 3 \,\log^2 \, 2 = 1.441359\ldots, $ and $ 2 + 1.441359\ldots > 3 = p_2. $
This is not known even under the Riemann hypothesis, which gives only $ \frac{p_{n+1}-p_n}{\sqrt{p_n}\log p_n}<\infty. $