# Does the sum of reciprocals of this subset of the primes converge?

Here

On the distribution of "nice" primes (primes $p$ , such that $\pi(p)$ is prime as well)

"nice" primes (primes $p$, such that $\pi(p)$ is prime as well) are introduced.

Is it known whether the sum of the reciprocals of the nice primes is convergent ?

The sum of the reciprocals of the primes in an arithmetic progression $an+b$ , where $a$ and $b$ are coprime and $n$ runs over the natural numbers, is divergent.

Intuitively, I would expect that there should exist enough nice primes to get a divergent series, but I may be wrong.

Has anyone an idea or a reference ?

• Another thing I am interested in : What is the largest known "nice" prime ? – Peter Jul 29 '16 at 23:18

Because of the Prime Number Theorem, ultimately we have $$.9\,n\,\log n \leq p_n \leq 1.1 \,n\,\log n$$ where $p_n$ is the $n$th prime.
The $n$th nice prime is $p_{p_n}$, so we ultimately get $$p_{p_n} \geq .9\,p_n\,\log(p_n)\geq .9 \,(.9\,n\,\log n)\,\log(.9\,n\,\log n).$$
Since that last logarithm is equivalent to $\log n$, we ultimately get $$p_{p_n} \geq .5\,n\,(\log n)^2$$ so $$\frac 1{p_{p_n}} \leq \frac{2}{n\,(\log n)^2} \qquad\text{ultimately,}$$ which proves that $\sum_n \frac{1}{p_{p_n}}$ converges, because $\sum \frac 1{n(\log n)^\alpha}$ converges iff $\alpha > 1$ (that's a Bertrand series, it's for example an application of Cauchy's condensation test or of an integral comparison).