Let $p_n$ be the $n$th prime number. Does the following series $$ \sum_{n \ge 1} \frac{1}{p_{p_n}} = \frac{1}{3} + \frac{1}{5} + \frac{1}{11} + \cdots $$ converge or diverge? Similarly, I am so curious about the convergence of the subsequent type of subserious, like $$ \sum_{n \ge 1} \frac{1}{p_{p_{p_n}}} $$, and so on.
Asked
Active
Viewed 110 times
2
-
I am sure it is a duplicate, $\sum_{n\geq 1}\frac{1}{p_{p_n}}$ is already convergent. – Jack D'Aurizio Dec 30 '14 at 22:44
-
3That just follows from proving that $p_{p_n}\gg n\log^2 n$, then applying Cauchy's condensation test. – Jack D'Aurizio Dec 30 '14 at 22:45
-
Oh, really! How about the next question? – hkju Dec 30 '14 at 22:47
-
1If $\frac{1}{p_{p_n}}$ is summable also $\frac{1}{p_{p_{p_n}}}$ is summable (it is a subsequence). – Jack D'Aurizio Dec 30 '14 at 22:49
-
How can we prove that $p_{p_n}\gg n\log^2 n$? – hkju Dec 31 '14 at 05:06
-
2From the PNT (or the Chebyshev's bound) we have $\pi(n)\ll\frac{n}{\log n}$, hence $p_n\gg n\log n$ and $p_{p_n}\gg n\log^2 n$. – Jack D'Aurizio Dec 31 '14 at 10:17