Lower bound for the 'de Polignac constant' Let's introduce the 'de Polignac constant'  $ K_{Pol} : =\sum_{i>0}2^{-k_{i}} $ , where  $ k_{i} $ is the  $ i $ -th positive integer such that  $ 2k_{i} $ is a Polignac number, i.e a number that is the difference of two consecutive primes in infinitely many ways. De Polignac's conjecture is equivalent to  $ K_{Pol}=1 $ .
Do we know a non trivial lower bound for  $ K_{Pol} $? 
Edit : a proof that  $ K_{Pol}>1/2 $ would entail the truth of the twin prime conjecture.

We can give much better estimates on $K_{Pol}$ than ones mentioned in Charles answer. All the results used are already proven in the same paper by Polymath, but instead of the results cited in the introduction, we can use stronger results encapsulated in Theorem 3.2, the relevant parts of which I will quote for completeness.
Definition Denote by $DHL[k,j]$ the statement "For any admissible $k$-tuple $\mathcal H=(h_1,\dots,h_k)$ there exist infinitely many translates $n+\mathcal H=(n+h_1,\dots,n+h_k)$ of $\mathcal H$ which contain at least $j$ primes".
Theorem 3.2 The following hold:


*

*$DHL[50,2]$, unconditionally.

*$DHL[3,2]$, under the assumption of GEH conjecture.


Also, this result isn't listed in the theorem, but on page 46 it is mentioned that EH implies $DHL[6,2]$, as proven by Goldston, Pintz and Yıldırım. I won't be doing full calculations in all cases, but I will prove an estimate using the result under GEH.
Claim Under GEH, either $2$ is a Polignac number, or, for every $n\geq 1$, either $6n$ or $6n-2$ is a Polignac number.
Proof The triple $(0,2,6n)$ is rather clearly admissible, so under GEH infinitely many translates of this triple contain two primes. If $2$ is not Polignac, then for each $n$, such pairs of primes must have difference $6n-2$ or $6n$. One of these must appear infinitely often.
Therefore either $K_{Pol}$ is at least $2^{-1}$, or at least $\sum_{n=1}^{\infty}2^{-3n}=\frac{1}{7}$. In either case, $K_{Pol}\geq\frac{1}{7}$. I don't think you can improve this bound, since every admissible triple will have a difference divisible by $3$, so it appears to be consistent that Polignac numbers are precisely the ones divisible by $6$.
