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Can anyone prove or disprove the following statement?

$$ \lim_{n \to \infty} \frac{p_{n+1} - p_{n}}{\sqrt{p_n}} = 0.$$

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I'm guessing $p_n$ is meant to the be $n$th prime? Ideally, one should not be forced to guess, though, and the government doesn't like it when I read minds without a warrant. –  Arturo Magidin Jul 23 '12 at 0:38
Partial results are available at en.wikipedia.org/wiki/Prime_gap#Further_results. –  sdcvvc Jul 23 '12 at 1:01

3 Answers 3

This is not known even under the Riemann hypothesis, which gives only $$ \frac{p_{n+1}-p_n}{\sqrt{p_n}\log p_n}<\infty. $$

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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 $$ 2 + 1.441359\ldots > 3 = p_2. $$

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Was "WookiePedia" intentional? –  Potato Jul 23 '12 at 2:17
@Potato, such a good feeling to have my work noticed. I work so hard for my fans. Also for everyone else. –  Will Jagy Jul 23 '12 at 2:20

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.

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The limit is $0$ if you change the exponent to $0.526$ –  Will Jagy Jul 23 '12 at 1:09

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