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I know that the sum of the reciprocals of the primes diverges, but I was asked of this question (title).

The problem stated that it will converge to a value, but I cannot figure out the value.

David Chen
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  • @SeanNemetz I have tried to 1. find some patterns to combine the elements to make them offset each other 2. find the lower bound and the higher bound and use the squeeze theorem – David Chen Mar 23 '18 at 03:48
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    Are you asking for the value of the sum, or only for a proof that the series converges? If the former (which I strongly doubt, but you are the one who knows), add this to the title. And anyway, please avoid in the future using noninformative titles. – Did Mar 23 '18 at 06:25
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    @Did I think my original title is pretty descriptive, and I clearly states that I want to figure out the value it converges to. The fact you cannot find the answer does not mean others cannot. – David Chen Mar 23 '18 at 06:31
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    The only version that was even remotely descriptive was mine. The present version does not even ask a question. Anyway, seeing your desire to be non constructive, I stop to try to help you. – Did Mar 23 '18 at 06:36
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    Why did you just change your title is the former one was "pretty descriptive"? Somebody is not making much sense here... – Did Mar 23 '18 at 06:38
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    @Did Maybe you can provide some reasons of why the original title is not informative. Besides, It is pretty rude to edit my question and basically change what it is asking. – David Chen Mar 23 '18 at 06:39
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    @Did You can at least point out why you think it is impossible to find the value it converges to, or you are just complaining but not helping. – David Chen Mar 23 '18 at 06:41
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    I did not change the body, only the title, which was abysmal. "Thank you" is not a forbidden word on this site, you know? – Did Mar 23 '18 at 06:41
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    Your string of recent comments is a strong deterrent to help you in any way, if you want to know. – Did Mar 23 '18 at 06:42
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    And now you accepted an answer which (unsurprisingly) does not compute the sum. Well well well... – Did Mar 23 '18 at 07:12

2 Answers2

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$\frac{1}{p(i)p(i+1)} < \frac{1}{p(i)^2}$, so $\sum \frac{1}{p(i)p(i+1)} < \sum \frac{1}{p(i)^2}$. On the other hand, $\sum \frac{1}{p(i)^2} < \sum \frac{1}{n^2}$, because the former sum is a subset of the latter. Finally, $\sum \frac{1}{n^2} = \frac{\pi^2}{6}$ is known to converge.

xyzzyz
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    The question asks what the value it will converge to. – Q the Platypus Mar 23 '18 at 04:28
  • @QthePlatypus I dont think it is even reasonable to try to comlpute the value of that sum. I guess the Op misprinted his question. I think the key point is to check the convergence. +1 – Guy Fsone Mar 23 '18 at 06:19
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Just for the fun of playing with "small" numbers.

Considering $$S_k=\sum_{i=1}^{10^k}\frac{1}{p_i\,p_{i+1}}$$ what I obtained (better say, my computer) is $$\left( \begin{array}{cc} k & S_k \\ 1 & 0.2944055709 \\ 2 & 0.3008401966 \\ 3 & 0.3010804833 \\ 4 & 0.3010924131 \\ 5 & 0.3010931251 \end{array} \right)$$ My computer (and I) gave up for $k=6$.

Inverse symbolic calculators do not seem to find anything.

The upper bound is $$\sum_{i=1}^{\infty}\frac{1}{p_i^2}=P(2)\approx 0.452247$$ where appears the prime zeta function.

Edit

You could be interested by this which gives $\approx 0.301093176358399894$

Claude Leibovici
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