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If $p$ is a prime number, and $q$ is its twin prime, the sum of the reciprocal twin numbers is convergent and the value of the sum of the series is the Brun constant. Now, if we consider the prime numbers $x=p+\alpha$ where $\alpha$ is a constant such that $x$ is also a prime, we can consider the sum: $$S=\frac{1}{\sum_{k=1}^{N}x_k}$$ The question is: is it possible to show for what values of $\alpha$ the series is convergent? Thanks in advance.

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$N$ is finite? So it always "converges", doesn't it? Maybe you should first prove that there are infintely many prime pairs with the given property...good luck. –  draks ... Jul 26 '12 at 9:04
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Brun's Constant for Cousin Primes $ B_4 = \left(\frac{1}{7} + \frac{1}{11}\right) + \left(\frac{1}{13} + \frac{1}{17}\right) + \left(\frac{1}{19} + \frac{1}{23}\right) + \cdots \approx 1.1970449 $ might be related... –  draks ... Jul 26 '12 at 9:24
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i think in this case,principle does not change much,we could have reciprocal sum of twin numbers + sum of inverse of (prime numbers+even numbers),i think result does not change much from Brun constant –  dato datuashvili Jul 26 '12 at 9:24
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@Riccardo.Alestra Your correction is still not correct. You are not adding any reciprocals together! –  Erick Wong Jul 26 '12 at 9:35
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Let me also ask you to edit the question. You have the undefined symbol $x_k$, the undefined symbol $N$, you refer to $S$ as a series when in fact it is the reciprocal of a finite sum; the question is badly broken. Please fix it. –  Gerry Myerson Jul 26 '12 at 12:12

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up vote 3 down vote accepted

For any non-zero integer $\alpha$, the Selberg sieve can be used to show that the number of primes $p \le n$ such that $p + \alpha$ is prime is at most $C_\alpha n/(\log^2 n)$. The value of $\alpha$ does have an effect: if $\alpha$ is divisible by many small primes, then the constant in the upper bound will be higher. On the other hand if $\alpha$ is odd, there is at most one prime satisfying the condition.

Either way, it is easy to show by partial summation that the reciprocal sum converges. Brun's original proof gave a slightly weaker upper bound, but one that is still strong enough to obtain convergence.

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