I've recently bumped into this old question of mine and I was wondering about the sum running over only numbers with distinct prime factors. More precisely:
For $k,n\ge1$ let $S_{k,n}$ be the set of numbers whose factorization contains exactly $k$ distinct primes out of the first $n$ ones. So, for instance: $S_{1,n}=\{2,3,...,p_n\};$ $S_{2,n}=\{6,10,...,2p_n,15,21,...,3p_n,...,p_{n-1}p_n\}.$
With both $k$ and $n$ allowed to vary, are the asymptotics of $$\sum_{m\in S_{k,n}} \frac1m$$ known? Up to what order?
For convenience I will use a slightly different notation.
Meissel-Mertens second theorem gives for the sum over primes $p \leq n$ that $$ \lim_{n \rightarrow \infty} \left( \sum_{p \leq n} \frac{1}{p} - \ln(\ln n)\right) = M$$
In other words asymptotically $$ \sum_{p \leq n} \frac{1}{p} \approx \ln(\ln n)$$
If we square the expression we get
$$ \left(\sum_{p \leq n} \frac{1}{p} \right)^2 = \sum_{p \leq n} \frac{1}{p^2} + \sum_{p_1 \neq p_2 \leq n} \frac{1}{p_1 p_2} \approx \left[\ln(\ln n)\right]^2 $$
Since the sum $\sum_p \frac{1}{p^2}$ converges, we therefor obtain $$ \sum_{p_1 < p_2 \leq n} \frac{1}{p_1 p_2} \approx \frac{1}{2} \left[\ln(\ln n)\right]^2 $$
Taking higher powers and retaining the leading terms yields $$ \sum_{p_1 < p_2 < \dots <p_k\leq n} \frac{1}{\prod_i^k p_i} \approx \frac{1}{k!} \left[\ln(\ln n)\right]^{~k} $$
Using the prime number theorem $p_n \sim n \ln n$ we find $$ \sum_{m \in S_{k,n}} \frac{1}{m} \approx \frac{1}{k!} \left[\ln(\ln (n \ln n))\right]^{~k} $$
Note that this only works here because you consider all combinations with factors up to a maximum prime. A different sum would be over all primes $p$,$q$ where the product is bounded $$ \sum_{p \neq q | p q \leq n} \frac{1}{p q}$$