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Does the series $\sum_{n=1 }^{\infty}1/p_{j} $ of reciprocal primes converge?

Experimentally, it seems convergent.

Yes
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    Look up Mertens' second theorem. It grows like $\log\log n$ – Empy2 Sep 08 '14 at 05:18
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    How can a sum of positive numbers go to $0$ when the sequence of partial sums is positive and increasing? – Adam Hughes Sep 08 '14 at 05:18
  • Do you mean the sequence of primes? If you can specify $(p_j)$ arbitrarily, you can certainly make this convergent. – Travis Willse Sep 08 '14 at 05:18
  • I mean the sequence of reciprocal primes. – Yes Sep 08 '14 at 05:19
  • Don't confuse sequences and series. Sequences are $a_1, a_2, a_3, \dots$. Series are the sum of these terms: $a_1 + a_2 + a_3 + \dots$. – MT_ Sep 08 '14 at 05:24
  • So it diverges according to Merten's second theorem? – Yes Sep 08 '14 at 05:24
  • @MichaelT: My gross typo, thanks. Was typing rushingly. :) – Yes Sep 08 '14 at 05:26
  • Yes, it diverges according to http://en.wikipedia.org/wiki/Mertens%27_theorems – Empy2 Sep 08 '14 at 05:29
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    Let $p_j$ be the $j$-th prime. The fact that $\sum_0^\infty \frac{1}{p_j}$ diverges was proved by Euler. The divergence is slow. Even a quite lengthy computation will give results that look consistent with convergence. – André Nicolas Sep 08 '14 at 05:33
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    Asking if it is convergent is reasonable. Asking if it tends to 0 is less reasonable. – almagest Sep 08 '14 at 06:24
  • @almagest: Then how does it have anything to with "off-topic"? Are you ... – Yes Sep 08 '14 at 06:35
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    I didn't say or imply it was off-topic. But it would have been good if you could have taken a little more trouble drafting your question! – almagest Sep 08 '14 at 06:38
  • @Comeseeconquer. +1 Thank you. – almagest Sep 08 '14 at 07:19
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    This is actually how Euler proves there are infinitely many primes. He does it by introducing what's now called the Riemann zeta function $\zeta(s)$, and using the fact that it has a pole at $s=1$ to show $\sum_p \frac{1}{p}$ diverges. If there were finitely many primes it would be a finite sum, and it would converge. This is the beginning of analytic number theory. – Zavosh Sep 08 '14 at 07:40
  • @PedroTamaroff: Yes, are you disagreeing with something I said? If you take the log of $\prod_p \frac{1}{1-p^{-s}}$ and expand $-\log (1-p^{-s}) = \frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots$, you can write $\log \zeta(s) = \sum_p \sum_n \frac{1}{p^{ns}}$. You split the double sum into $\sum_p \frac{1}{p^s}$ plus everything else, and you show everything else converges, so $\sum_p \frac{1}{p^s}$ diverges as $s \rightarrow 1^+$, because $\zeta(s)$ does. Therefore the sum $\sum_p \frac{1}{p^s}$ must be infinite. This is the form I've always seen the argument in when the details were provided. – Zavosh Dec 05 '14 at 08:05
  • Ok, granted. ${}{}{}{}{}{}$ – Pedro Dec 05 '14 at 09:35
  • This question has been asked before: http://math.stackexchange.com/questions/15946/does-the-sum-of-reciprocals-of-primes-converge and http://math.stackexchange.com/questions/674877/what-is-the-sum-of-the-reciprocal-of-primes-yes-it-diverges – robjohn Dec 05 '14 at 14:25

2 Answers2

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Here is an easy proof of the divergence of the series of prime reciprocals, which I saw in the American Mathematical Monthly in a bygone millennium. Assume for a contradiction that the series converges. Choose $N$ so that $$\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\lt1.$$ Then $$S=1+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)^2+\cdots$$ is a convergent geometric series.

Let $d=P_1P_2\cdots P_N$ and consider the series $$\frac11+\frac1{1+d}+\frac1{1+2d}+\frac1{1+3d}+\cdots$$ which of course diverges. However, each denominator $1+nd$, since it is not divisible by any of the first $N$ primes (note the resemblance to Euclid's proof), is a product of primes $\ge P_{N+1}$. Thus each term $\frac1{1+nd}$ occurs in the geometric series $S$ when the powers are expanded, that is, $$\frac11+\frac1{1+d}+\frac1{1+2d}+\cdots\lt1+\left(\frac1{P_{N+1}}+\frac1{P_{N+2}}+\cdots\right)+\left(\frac1{P_{N+1}}+\cdots\right)^2+\cdots\lt\infty$$ and this contradiction proves the theorem.

If anyone reading this knows the source of this proof, please edit the attribution into my answer. Thanks.

The proof is given by Clarkson [Clarkson, James A. ( 1966) On the series of prime reciprocals. Proc. Amer. Math. Soc .. /7: 541; MR 32. #5573.] https://www.ams.org/journals/proc/1966-017-02/S0002-9939-1966-0188132-7/S0002-9939-1966-0188132-7.pdf

Riemann
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In this answer, it is shown that $$ \sum_{\substack{p\le n\\p\text{ prime}}}\frac1p=\log(\log(n))+O(1) $$ and therefore, the sum diverges.

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robjohn
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  • would the downvoter care to comment? This seems to answer the question, so would I think it's somewhat helpful. I referenced the other answer so that I would not need to duplicate the other answer here. – robjohn Dec 07 '14 at 18:14