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This question already has an answer here:

Surprised that nowhere in web or in MSE is asked it and here is the not considering all of the composite numbers.

Does $$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+ \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \dots$$ converge? If so, how?

Both the Harmonic series and the sum of the reciprocals of all the prime numbers do not converge which doesn't help to answer the question.

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marked as duplicate by Bumblebee, John B, Ethan Bolker, Alex M., Fly by Night Apr 19 '16 at 16:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The primes have zero natural density so... $\endgroup$ – PITTALUGA Apr 8 '16 at 15:58
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    $\begingroup$ You have a lower bound given by the sum of all reciprocals of even numbers bigger than 2, which is essentially half of the harmonic series. So it must diverge. $\endgroup$ – Brandon Carter Apr 8 '16 at 15:58
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    $\begingroup$ math.stackexchange.com/questions/1211498/… $\endgroup$ – Bumblebee Apr 19 '16 at 6:12
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Simple answer : For all prime $p$ (except for 2 and 3), $p-1$ is composite, so $\sum_{p\in\mathcal P-\{2,3\}}\frac 1 p < \sum_{p\in\mathcal P-\{2,3\}}\frac 1 {p-1} < \sum_{n\notin\mathcal P}\frac 1 n$

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Since $\sum_{n=1}^\infty \frac{1}{n}$ diverges, so does $\sum_{n=2}^\infty\frac{1}{2n}$. All the terms in this series are included in your series, so your series also diverges.

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    $\begingroup$ I like this answer way better than the accepted one - it's much more intuitive IMO. $\endgroup$ – fluffy Apr 8 '16 at 21:30
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    $\begingroup$ I think it is important to mention that this property (that adding extra terms cannot make a divergent series converge) is only true for series where each term is positive. Otherwise the alternating harmonic series is an easy counterexample. $\endgroup$ – Fengyang Wang Apr 9 '16 at 4:11
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    $\begingroup$ @FengyangWang: I disagree -- I think this answer is exactly as long as it needs to be. $\endgroup$ – TonyK Apr 9 '16 at 14:31
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    $\begingroup$ @fluffy - This answer is very nice indeed (and +1). The accepted answer is a bit more 'romantic' ;) $\endgroup$ – user231343 Apr 9 '16 at 20:14
  • $\begingroup$ I like this answer, the sum of reciprocals of primes is a big jackhammer $\endgroup$ – enthdegree Apr 14 '16 at 8:49
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This sum does not converge. Consider for example the sum of the reciprocals of all the even numbers greater than 2: $1/4 + 1/6 + 1/8 + 1/10 + \ldots$. This diverges, the proof being analogous to the proof that the harmonic series diverges. Thus your series has arbitrarily large partial sums and thus diverges.

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If $p_n$ is the $n$th prime number,

Note $$(\sum_{i=1}^{n}\frac{1}{p_i})^2(\text{Divergent})=\sum_{i=1}^{n}\frac{1}{(p_i)^2}(\text{Convergent})+2\sum_{1 \le i < j \le n}^{n}\frac{1}{p_ip_j}$$ Is true.

We have that $$\sum_{1 \le i \le j \le n}^{n}\frac{1}{p_ip_j}$$Diverges. But all numbers that are a multiple of two primes are composite.

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