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Let $P_i$ be sequence of prime numbers i.e $P_1=2,P_2=3,P_3=5 ...$

Euler has proved that the sum $$\sum_{i=1}^{\infty}\dfrac{1}{P_i}$$ is divergent.

Set $a_i=P_{P_i}$ then $a_1=P_2=3$ , $a_2=P_{3}=5$ and $a_3=P_5=11$ ...

My question is that $$\sum_{i=1}^{\infty}\dfrac{1}{a_i}$$ is convergent or not. From a more general perspextive if set $$a_i=P_{P_{P_{._{._{P_i}}}}}$$ (n times sub) then can we conclude that

$$\sum_{i=1}^{\infty}\dfrac{1}{a_i}$$ is convergent ? if it is so, what is the minumum of such $n$ ?

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marked as duplicate by daniel, Hagen von Eitzen, Joel Reyes Noche, Mark Bennet, Milo Brandt Dec 4 '14 at 23:11

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    $\begingroup$ This is a dupe but I can't find it at the moment. It's convergent. $\endgroup$ – daniel Dec 4 '14 at 18:41
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If you manage to prove $$ p_n \geq (1-\varepsilon)\, n \log n \tag{1}$$ for a certain $\varepsilon \geq 0$, for any $n$ big enough ($n\geq n_0$), then: $$ \sum_{n\geq n_0}\frac{1}{p_{p_n}} \leq \frac{1}{1-\varepsilon}\sum_{n\geq n_0}\frac{1}{p_n \log p_n} \leq \frac{1}{(1-\varepsilon)^2}\sum_{n\geq n_0}\frac{1}{n \log n\log p_n}\leq \frac{1}{(1-\varepsilon)^2}\sum_{n\geq n_0}\frac{1}{n \log^2 n}$$ is convergent since $$ \sum_{n\geq 2}\frac{1}{n\log^2 n}$$ is convergent by Cauchy's condensation test. In order to prove $(1)$, you may use the PNT or the Chebyshev's weaker version.

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  • $\begingroup$ Or just use Rosser's theorem and see that $p_n > n\ln n$ :-) $\endgroup$ – Peter Košinár Dec 4 '14 at 18:57
  • $\begingroup$ @PeterKošinár: I wasn't aware of that result. It is really nice, it gives that we can just take $\varepsilon=0$, many thanks. $\endgroup$ – Jack D'Aurizio Dec 4 '14 at 18:59

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