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I would like that all of you interested in number theory and specially, in prime numbers, to take a look at this quite elementary approach on the problem related to the conjecture of Polignac, which says that for every $k \in \mathbb N$ there exist infinitely many consecutive prime numbers with the difference $2k$ (case $k=1$ is the twin primes conjecture).

Let us suppose that for every $k \in \mathbb N$ the set $S_k=\{n\in\mathbb N|\;(p_{n+1}-p_n=2k)\}$ is finite (which could also be the case). This means that for every $k\in\mathbb N$ there exist $a(k)\in\mathbb N$ such that $p_{a(k)+m+1}-p_{a(k)+m}>2k$, for every $m\in\mathbb N$. But this implies that $\ lim_{n\to\infty} (p_{n+1}-p_n)=\infty$.

So this observation says that if $lim_{n\to\infty}(p_{n+1}-p_n)\neq\infty$ then there exist at least one $k\in\mathbb N$ such that $S_k$ is infinite, and there is connection with conjecture of Polignac.

It could be the case that somewhere in this short proof I made the mistake so I welcome comments about the proof itself, do you see any error?

And can we show somehow that $lim_{n\to\infty}(p_{n+1}-p_n)\neq\infty$?

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Could you please explain how you deduced your inequality from the assumption that $S_k$ is finite for any positive integer? Thanks. –  awllower Jan 31 '13 at 10:30
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So far as I know, no one has been able to disprove $\lim_{n\to\infty}(p_{n+1}-p_n)=\infty$. –  Gerry Myerson Jan 31 '13 at 10:35
    
@awllower : If $S_k$ is finite then all the natural numbers greater then the maximal $n\in\ S_k$, in other words all such $m>n$ must imply $p_{m+1}-p_m>2k$, it is obvious to me. –  A.P. Jan 31 '13 at 10:42
    
Gerry, what about the proof, do you agree with all the steps? –  A.P. Jan 31 '13 at 10:42
    
What if $p_{m+1}-p_m<2k$? –  awllower Jan 31 '13 at 10:43

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