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.
Convergence of a series of reciprocal prime numbers
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elementary-number-theory
prime-numbers
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1$N$ 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. – 2012-07-26
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0@draks:Obviously $N$ is finite. – 2012-07-26
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0@Riccardo.Alestra What is exactly is $N$, then? – 2012-07-26
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0Sorry. N is obviously infinite – 2012-07-26
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0I consider the limit for $N \to \infty$ – 2012-07-26
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0we should consider such situation if $P$ is prime and $x=p+a$ is also prime then we may have twin primes at first sum and then $a$ could be even,so we have sum of prime numbers+sum of even numbers – 2012-07-26
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0sure it depends on $N$ – 2012-07-26
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0Sorry. I corrected the question – 2012-07-26
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1[Brun's Constant for Cousin Primes](http://en.wikipedia.org/wiki/Cousin_prime) $ 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... – 2012-07-26
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1i 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 – 2012-07-26
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1@Riccardo.Alestra Your correction is still not correct. You are not adding any reciprocals together! – 2012-07-26
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1Let 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. – 2012-07-26
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1@Riccardo.Alestra can you edit your formula or the title/text please? It just doesn't fit together... – 2012-07-26
1 Answers
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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.