A series with the non square free numbers $q_n:\:$ $\sum_{k=1}^\infty\frac{(-1)^{1+\Omega(q_k)}}{q_k}=0$ Let $q_n$ be the $n$-th natural number that can be divided by a square $>1$ (https://oeis.org/A013929). Conjecture: 
$$\sum_{k=1}^\infty\frac{(-1)^{1+\Omega(q_k)}}{q_k}=0$$
where $\Omega(n)$ is the number of (not necessarily different) prime factors of $n$. 
I'm sure that the conjecture is true, but I can't prove it and that's what I need help with.
The conjecture is related with An accountancy of the natural numbers. The corresponding series for squarefree numbers $s_n$
$$\sum_{k=2}^\infty\frac{(-1)^{1+\Omega(s_k)}}{s_k}=1$$

This must be due to the fact that the reciprocals of all squarefree numbers with whatever signs appear in the $\prod\left(1\pm{1\over p}\right)$, and the reciprocals of all natural numbers appear in another product, $\prod{1\over1\pm{1/p}}$, and both products are easy to evaluate.
Indeed,
$$\sum_{n=1}^\infty{(-1)^{\Omega(n)}\over n}=\prod_{prime\;p}\left(1-{1\over p}+{1\over p^2}-{1\over p^3}+\dots\right)={1\over\prod\left(1+{1\over p}\right)}=0$$
On the other hand, look at the similar sum over squarefree numbers:
$$\sum_{squarefree\;s}{(-1)^{\Omega(s)}\over s}=\prod_{prime\;p}\left(1-{1\over p}\right)=0$$
Your sum over squareful numbers is just the difference between the two.
