TITLE: Define an infinite subset of primes such that the sum of reciprocals converges
QUESTION [12 upvotes]: 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?

REPLY [6 votes]: I see lots of answers giving subsets of primes yielding different upper bounds on $\sum{\frac{1}{p_n}}$, but none stating explicitly what the bound is. So I thought I'd give a subset for which the exact sum is known.
Let $x$ be any positive real number. Now define $p_n$ to be the smallest prime such that $\sum_{i=1}^{n}{\frac{1}{p_i}} < x$ and $\forall i<n: p_n \neq p_i$. Then it holds that $\sum_{i=1}^{\infty}{\frac{1}{p_i}} = x$