Do we know the rate of divergence of the sum of reciprocals of the $k$-almost primes?

Question

A $k$-almost prime is a positive integer having exactly $k$ prime factors, not necessarily distinct. Let $\mathbb{P}_k$ be the set of the $k$-almost primes and let $$ \rho_k(n):=\sum\limits_{\substack{q\in \mathbb{P_k}\\q\le n}}\frac1q. $$ So with $k=1$ and $n\to\infty$ we have the prime harmonic series, which diverges like $\log\log n+ B_1$, $B_1$ being the Meissel-Mertens the constant.. $\rho_2(n)$ diverges at a rate of $\frac{\log^2\log n}{2}+\frac15$. In general, is the asymptotical behaviour of $\rho_k(n)$ known?

Answer 1

The first-order asymptotics are easy. The counting function of $\mathbb P_k$ satisfies the Poisson-like asymptotic:

$$P_k(x) := \#\{n \le x : n \in \mathbb P_k\} \sim \frac{x}{\log x}\cdot \frac{ (\log \log x)^{k-1}}{(k-1)!}.$$

By partial summation, the reciprocal sum is $$ \sum_{\substack{n \in \mathbb P_k \\ n \le x}} \frac{1}{n} = \int_{t=2}^x \frac{d P_k(t)}{t} = \frac{P_k(t)}{t} \bigg|_{t=2}^x + \int_{t=2}^x \frac{P_k(t)}{t^2} dt.$$

The first term on the right is $O(1)$ since $P_k(x)/x \to 0$, so all the divergence in the sum is coming from the integral term, which is asymptotic to:

$$ \frac{1}{(k-1)!}\int_{t=2}^x \frac{(\log \log x)^{k-1}}{t \log t} dt = \frac{1}{(k-1)!} \int_{u=\log \log 2}^{\log \log x} u^{k-1} du = \frac{1}{k!}(\log \log x)^k + O(1).$$

Edit: The above is asymptotic to the reciprocal sum, which is not to say that $O(1)$ is the correct error term. In fact, the asymptotic for $P_k(x)$ has a relative error of $O(1/\log \log x)$ (the error term must depend in some way on the specific definition of $k$-almost prime, like whether duplicates are allowed), so the same is true of the estimate above. I would expect an error term of size $O(\log \log x)^{k-1}$.