Is the series $\sum 1/p$ in which p is prime converges? I remembered there's some theorem about this. THank you
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As an interesting exercise, try to come up with an infinite subset of primes, such that the sum of the reciprocal elements of that subset converges. You might be surprised of how non-trivial this task is. – barak manos Nov 21 '16 at 16:29
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@barakmanos: $a_0=2$ and $a_n={}$smallest prime${} \ge 2a_{n-1}$? – hmakholm left over Monica Nov 21 '16 at 16:31
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1@HenningMakholm: What??? – barak manos Nov 21 '16 at 16:31
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1It diverges because $\log \zeta(s) = \sum_p p^{-s} + \sum_{k=2}^\infty \frac{p^{-sk}}{k}$ where the second series converges for $Re(s) > 1/2$, and $\zeta(s)$ has a pole at $s=1$ therefore $\lim_{s \to 1^+} \sum_p p^{-s} = + \infty$ so $\sum_p p^{-1}$ cannot converge (series of positive terms) – reuns Nov 21 '16 at 16:32
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@barakmanos: That generates an infinite sequence of primes such that the sum of their reciprocals converges. – hmakholm left over Monica Nov 21 '16 at 16:32
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@HenningMakholm: Not sure I understand the term "generates" in this context (I see only two elements here - $a_0$ and $a_1$). – barak manos Nov 21 '16 at 16:33
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@barakmanos $\sum_n \frac{1}{p_{n^2}}$ converges without problems, of course it is not easy to find if $q \in { p_{n^2}}$ so maybe that's what you meant – reuns Nov 21 '16 at 16:33
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@user1952009: Let alone proving this claim. – barak manos Nov 21 '16 at 16:34
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@HenningMakholm: Hahaha. You've used the convergence of $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}$... nice :) – barak manos Nov 21 '16 at 16:36
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@HenningMakholm: I posted that as a question quite a while ago. The best answer there is somewhat similar to yours now that I think about it, but if you think it's different enough then please feel free to add it there... – barak manos Nov 21 '16 at 16:38
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The answer is no. You can consider the function $$ P(z)=\sum_{p=prime}\frac{1}{p^z}$$ and you can show that this series converges for $\Re(z)>1$ . One can also prove that the line $\Re(z)=0 $ is a natural boundary for the function we defined eariler, which means that $P(z)$ has no analytic continuation on the left half plane.
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