Sum of the reciprocal of the prime-position primes. The primes are $2, 3, 5, 7, 11, 13...$
The sum of the reciprocals of the primes diverges, proven by Euler:
$$\sum_{n=1}^\infty{\frac{1}{p_n}}=\infty$$
Here, $p_n$ is the $n$-th prime.
I'm asked to prove whether the following converges or diverges:
$$\sum_{n=1}^\infty{\frac{1}{p_{p_n}}} = \frac{1}{3} + \frac{1}{5} + \frac{1}{11} + \cdots$$
I think that it diverges, but I don't know how to show it. Is it true that if I take any infinite subset of a diverging series, the subseries will diverge? That doesn't sound true.
$$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$$
The above series is a subseries of $(1 + \frac{1}{2} + 1 + \frac{1}{4} + 1 + \frac{1}{8} + \cdots)$. The full series diverges but the subseries converges, and both are infinite in length.
How do I solve this problem?

I think it is a duplicate, but in any case, if we have a weak version of the PNT, stating that $\pi(n)\leq\frac{Cn}{\log n}$ (for instance, Chebyshev's bound), we also have $p_n\geq D\,n \log n$ and $p_{p_n}\geq D\, n \log^2 n$ for any $n$ big enough. 
The series $\sum_{n\geq 2}\frac{1}{n\log^2 n}$ is convergent by the Cauchy's condensation test, hence $\sum_{n\geq 1}\frac{1}{p_{p_n}}$ is convergent, too.
