Does there exist an approximate formula which can calculate the sum of the reciprocal of squarefree number which have an even number of prime factors? Example :
The sum of the reciprocal of the squarefree numbers $\text{Se}(X)$ with an even number $2k$ of prime factors, where $k>0$
For $k$ less than $49$:
$\text{Se}(49)=\frac{1}{6}+\frac{1}{10}+\frac{1}{14}+\frac{1}{22}+\frac{1}{26}+\frac{1}{34}+\frac{1}{38}+\frac{1}{46}+\frac{1}{15}+\frac{1}{21}+\frac{1}{33}+\frac{1}{39}+\frac{1}{35} = \frac{77 \ 890 \ 556}{111 \ 546 \ 435}$
I want an approximate formula for $\text{Se}(X)$.
 A: Some hopefully useful information and heuristic ideas here, followed by empirical conclusions. Note that I'm writing on the assumption that you're actually interested specifically in semiprimes (i.e., primes with exactly two distinct prime factors). Including all squarefree $2k$-almost primes seems a bit odd, as the different layers of $k$-almost primes will surely have different properties.
Edit for clarity: This answer assumes we are working only with semiprimes, or $2$-almost primes. Specifically, the reciprocals are of the set
$$\{n \in \Bbb{N} \mid \omega(n) = \Omega(n) =2\}$$
where $\omega(n)$ is the number of distinct prime factors of $n$, and $\Omega(n)$ is the total number of prime factors of $n$.

Landau showed that the number of squarefree semiprimes less than or equal to $n$ is
$$\pi_{ss}(n) \approx \frac{n \log \log n}{\log n}$$
Interestingly, it's a very close parallel to the prime number theorem, and allows us to say that the $n^{\text{th}}$ squarefree semiprime is $s_{s,n} \approx n \log n / \log \log n$, and the probability of $n$ being a squarefree semiprime is $P(\omega(n) = \Omega(n) = 2) \approx \log \log n / \log n$. So we should expect to find about $\log \log n$ times as many semiprimes below $n$ as there are primes below $n$.
Separately, from Mertens's Second Theorem we have:
$$\sum_{p \le n} \frac1p \sim \log \log n + M$$
where $M \approx 0.26$ is the Meissel-Mertens constant. Surely, then, we could expect our desired sum to be, say, the above times another factor of $\log \log n$? Well... not so fast. First we should note that while there might be many more semiprimes than primes below $n$, the sum of the prime reciprocals gets a really nice headstart. After all, the first three prime reciprocals sum to just over $1$, but the first three semiprime reciprocals sum to not quite $\frac13$. The sum of semiprime reciprocals doesn't overtake the sum of prime reciprocals until around $n=80000$.
So, while $\log \log n <1$ up to $n=15$, our multiplier needs to grow a bit more slowly--something close to $1$ at $n=80000$ or so. If we decide to look empirically at the data, it grows about halfway between $\log \log n$ and $\log \log \log n$. If we let a curve fitter take on the data, we find that the multiplier we need is very close to:
$$\gamma \log \log \sqrt{n}$$
where $\gamma \approx 0.5772157$ is the Euler-Mascheroni constant, which happens to appear in Mertens's Second Theorem. (Frankly, when you're discussing something related to the primes, and the curve-fitting software spits out $\approx 0.5780$ with no tweaking, what are the chances that's just $\gamma$ hiding in the details? I'm gonna say high enough.)
So: empirically, looking only below $10^9$ so far, your sum seems to be within a hair of:
$$\sum_{k \le n}^{\omega(k) = \Omega(k)=2} \frac1k \approx \left(\log \log n + M \right)\left(\gamma \log \log \sqrt{n} \right) + O(1)$$
Note that both approximations take a while to get close to the correct values; these approximations are at best inaccurate below $n=100$, and are at best "good" below $n=10000$.

Note also, of course, this is not a proven result; possibly someone with better understanding of number theory can get you that. This is entirely empirical, but it comes close enough (and makes enough sense) that I'm willing to call it, at least, useful.
