It is known that
$$\sum_{k=1}^n \frac1k \sim \ln(n)$$
and
$$\sum_{p \text{ prime}}^n \frac1p \sim \ln(\ln(n)),$$
with the property that the difference in each case converges to a constant (Euler–Mascheroni and Meissel–Mertens constant, respectively).
Let's write $A_1 = \mathbb N$ and $A_2 = \mathbb P$ (the set of prime numbers). Can this principle be continued in the sense that there is an interesting description of a subset $A_3 \subset \mathbb N$ (or maybe even $A_3 \subset A_2$) such that
$$\sum_{k \in A_3}^n \frac1k \sim \ln(\ln(\ln(n)))$$
or, in general, $A_i \subset \mathbb N$ (or maybe even $A_i \subset A_{i-1}$) such that
$$\sum_{k \in A_i}^n \frac1k \sim \ln^{i}(n), \quad \text{where }\ln^i = \underbrace{\ln \circ \dots \circ \ln}_{i\text{ times}},$$
each with the property that the respective difference converges to a constant?
I remember a number theory book that I can't find now stating that the sum of the inverses of numbers of the form $n \log n$ diverges as $\log \log n$. The primes are very much like those because of the prime number theorem. It went on to say that the sum of the inverses of the numbers of the form $n \log n \log \log n$ diverges as $\log \log \log n$ and the pattern continues. It said that if you raise the last term to the $1+\epsilon$ power in any of these the sum converges. This is a quite natural progression.
No proof was given, but converting the sum to an integral makes it make sense. $$\int \frac {dx}{x \log x \log \log x}= \log \log \log x +c$$ for example.