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It is well known that the series $\sum_{p\in \mathbb{P}} \frac{1}{p}$ diverges where $\mathbb{P}$ denotes the set of primes. Brun proved that $\sum_{p\in \mathbb{P_2}} \frac{1}{p}$ converges where $ \mathbb{P_2}$ denotes the set of twin primes. Now for an even integer $k$ let $\mathbb{P}_k = \{ p, q \in \mathbb{P}: |p-q| \leq k\}$. By Zhang's result we know that $\mathbb{P}_k$ is infinite for $k$ larger than some threshold $N$. My question is what is the smallest value of $k$ for which $\sum_{p\in \mathbb{P}_k} \frac{1}{p} = \infty$? Or does no such finite value exist? Moreover, what are the asymptotics for these sums as $k \to \infty$?

Edit (S.K.): Since the question has more or less been answered in the comments and has already 2 close votes, let me try to turn this into a more on-topic question: does the series still converge if we take the sum over all reciprocals of primes $p$ whose distance to the next prime is less than $\frac{\log p}{\log \log p}$?

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    $\begingroup$ So the crudest heuristics suggest that this should be finite for all $k$: if each number $n$ is `prime with probability $1/\log n$', then the probability $n$ belongs to $\mathbb P_k$ is around $2k/(\log n)^2$. Hence you should expect $\sum_{p\in\mathbb P_k}1/p\approx \sum_{n>1} 2k/(n(\log n)^2)<\infty$, and expect that this should grow like $k$. More refined heuristics take into account small factors of $n$ etc. I doubt this would change the expected outcome much. $\endgroup$
    – Anthony Quas
    Commented Oct 20, 2014 at 23:14
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    $\begingroup$ One might think that Brun's argument generalizes to arbitrary $k.$ $\endgroup$
    – Igor Rivin
    Commented Oct 20, 2014 at 23:49

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