Define an infinite subset of primes such that the sum of reciprocals converges

How can we define an infinite subset of primes such that the sum of reciprocals converges?

$S=\{p\in \mathbb{Z}^+ : p\ \text{is prime and some condition on}\ p\}$ s.t. $\sum\limits_{p\in{S}}\frac{1}{p}\neq\infty$

A few options that come to mind for the condition on $p$ are:

• $\log_2(p+1)\in\mathbb{N}$
• $\log_2(p-1)\in\mathbb{N}$

But it has not been proved that there are infinite many such primes for either one of these options.

Any ideas?

For every $n\in \mathbb{N}$ define prime $p_n$ as the smallest of the primes which are greater than $2^n$. Then the set $\{p_n \mid n\in \mathbb{N}\}$ is infinite and

$$\sum_n \frac{1}{p_n} \leq \sum_n \frac{1}{2^n} = 1\text{.}$$

Let $p_n$ be a prime divisor of $(n^2)! + 1$. Then we have $p_n > n^2$, which means that $$\sum_{n \geq 1} \frac{1}{p_n} < \sum_{n \geq 1} \frac{1}{n^2} = \frac{\pi^2}{6} < \infty.$$