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That is, let $p_n$ be the nth positive prime number. Does $$L = \lim\limits_{n \to \infty} \left( p_{n+1} - p_n \right)$$ equal infinity?

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I doubt whether the behavior of that sequence is completely known, since it is still unsolved whether there are infinitely many twin primes. –  Tobias Kildetoft May 6 '11 at 7:47

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up vote 22 down vote accepted

No, the limit (probably) does not exist.

The sequence $p_{n+1} - p_n$ has a name: it is called the sequence of prime gaps. Define $g_n = p_{n+1} - p_n$, then you are interested in the sequence $g_1, g_2, \dots$. The following facts are known:

  • For any $k$, it is easy to see that the $(k-1)$ numbers $k!+2, k!+3, \dots, k!+k$ are all non-prime. Thus there exist arbitrarily long sequences of composite numbers; in other words, $g_n$ can get arbitrarily large: for any $k$ there exists $n$ such that $g_n \ge k$. Equivalently, $$ \limsup_{n \to \infty} (p_{n+1} - p_n) = \infty.$$

  • There is the (believed) twin-prime conjecture which states that there exist infinitely many primes that differ by 2; this means that $g_n$ takes the value $2$ for infinitely many $n$, or $$\liminf_{n \to \infty} (p_{n+1} - p_n) = 2.$$ Even if it turns out to be false for 2, there is Polignac's conjecture that for any even integer $k$, there exist infinitely many primes that differ by $k$; this has not been proved or disproved for any $k$. Even better, it has been proved in 2005, assuming a certain conjecture (which I think is weaker than the Riemann hypothesis) that there are infinitely many $n$ for which $g_n$ is at most $16$, thus $$\liminf_{n \to \infty} (p_{n+1} - p_n) \le 16.$$

  • The "average" gap between the $n$th prime and the next increases logarithmically: $g_n \approx \ln p_n$. So you can say that the distance between primes does become infinite "on average"; though infinitely often it touches very small numbers.

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I accepted your answer essentially because of the last topic. I should have thought in terms of average too, quite interesting. –  Luke May 7 '11 at 21:37
    
I just found this survey article that has a lot more details: ams.org/journals/bull/2007-44-01/S0273-0979-06-01142-6/… ("Small gaps between prime numbers: The work of Goldston-Pintz-Yildirim") –  ShreevatsaR May 15 '13 at 6:06
    
Polignac turned out to be right: golem.ph.utexas.edu/category/2013/05/…. –  Luke May 20 '13 at 23:48
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@Luke: AFAICT, Zhang's result does not yet prove Polignac's conjecture. It says that there are infinitely many prime gaps less than $C \approx 7 \times 10^7$, so technically all we can say is that at least one gap less than $C$ occurs infinitely often (not all gaps). But, it's a step in the relevant direction. –  ShreevatsaR May 22 '13 at 6:40

Assuming some fairly reasonable conjectures, then $(p_{n+1}-p_n)$ is a divergent sequence (i.e. a limit does not exist). For example:

Conjecture: There exists an infinite number of twin primes. (where $p_{n+1}-p_n=2$ an infinite number of times)

Conjecture: There exists an infinite number of prime pairs that differ by 4. (where $p_{n+1}-p_n=4$ an infinite number of times)

However, we can observe that $\lim\sup (p_{n+1}-p_n)=\infty$, since $n!+k$ is composite (properly divisible by k) for all $n \geq 2$ and $2 \leq k \leq n$. Therefore there are arbitrarily large prime gaps.

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If the limit is not infinity, then there can be found infinitely many primes differing by less than some bound M. Answering whether that is the case seems to be an open problem still.

The closest (contingent) answer after some quick searching seems to be the one noted at the bottom of this page about the Elliott–Halberstam conjecture stating that there are infinitely many pairs of primes differing by at most 16 (they used the linked conjecture in their proof).

If that were true, then the answer to your question would be no.

Though, maybe it is of interest to you that there are arbitrarily large gaps between primes, which is a related question. The proof is pretty simple: consider the numbers following $n!$...

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The numbers from $n!+2$ through to $n!+n$ are clearly all composite, though these sized sequences of $n-1$ consecutive composite numbers in fact appear a lot earlier for $n>3$. –  Henry May 11 '11 at 1:17

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