An accountancy of the natural numbers I guess this formula is known since I believe it's true 
$$\sum_{k=2}^\infty\frac{(-1)^{1+\Omega(k)}}{k}=1$$
where $\Omega(k)$ is the number of prime factors of $k$ (not necessary different primes). I can't find it and want a formal proof or a reference.
My intuition about this sum is something like this:

The probability of a number to have the prime factor $p$ is
  $\frac{1}{p}$ and to have the prime factors $p$ or $q$ is
  $\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}$ etc. The probability of a number $>1$ to have a prime factor is $1$.

Is this sum really not known? Haven't anything been published about it before?

I finally realize that the series
$$\sum_{k=1}^\infty\frac{(-1)^{\Omega(k)}}{k}=0$$
has been known for a long time, and that solves my confusion.

 A: The result is a simple consequence of the properties of Dirichlet series. 
Since $f(n)=(-1)^{\Omega(n)}$ is a multiplicative function, for any $s>1$ we have
$$\sum_{n\geq 1}\frac{f(n)}{n^s} = \prod_{p}\left(1-\frac{1}{p^s}+\frac{1}{p^{2s}}-\frac{1}{p^{3s}}+\ldots\right)=\prod_p\left(1+\frac{1}{p^{s}}\right)^{-1}=\frac{\zeta(2s)}{\zeta(s)} \tag{1}$$
by Euler's product. Since $\zeta(s)$ has a simple pole at $s=1$ with residue $1$, by taking the limit as $s\to 1^+$ we get
$$\boxed{ \sum_{n\geq 2}\frac{(-1)^{1+\Omega(n)}}{n}=\color{red}{1} }\tag{2}$$
as wanted. As pointed out by Kcd in the comments, we switched a sum and a limit to state the equality above, i.e. we exploited some form of the dominated convergence theorem, that in this case is given by the principle of analytic continuation. So, to the able to state $(2)$ with full rigor, we have to prove first that
$$ \sum_{n\leq N}(-1)^{\Omega(n)}=o(N) \tag{3}$$
holds. But that is a standard result in sieve theory. For instance, we may go through the following lines: we may choose at first some prime $q$ on the interval $[2,n]$ such that the vast majority of integers in such interval are $q$-smooth. Then the LHS of $(3)$, restricted to such integers, is not too difficult to approximate, and with an accurate choice of $q$ we may simply neglect the contribute given by the other (non-smooth) integers. An alternative is to notice that $(-1)^{\Omega(n)}$ equals $\mu(n)$ if $n$ is a square-free number, square-free numbers have a positive density among integers ($\frac{6}{\pi^2}$) and $\sum_{n\leq N}\mu(n)$ is well studied, so we may try to separate terms appearing in the LHS of $(3)$ according to their square-free part or their largest squarefree divisor and apply double-counting. Luckily, we don't have to be extremely accurate to prove $(3)$, since an almost ridiculous upper bound like
$$ \sum_{n\leq N}(-1)^{\Omega(n)}\leq\frac{N}{\sqrt{\log\log\log N}}$$
is yet good enough. Moreover, at first sight it looks that both
$$ \sum_{n\leq N}\mu(n)=o(N),\qquad \sum_{n\leq N}(-1)^{\Omega(n)}=o(N)$$
are equivalent to $\zeta(s)\neq 0$ over $\text{Re}(s)=1$, but I may be wrong about that.
A: We have
$$\sum_{k=2}^{\infty} \frac{(-1)^{1 + \Omega(k)}}{k} = 1 -\prod_p \left(1 - \frac{1}{p} + \frac{1}{p^2} \mp\right) = 1 -\prod_{p} \frac{1}{1 + 1/p}$$
The RHS goes to $1$ since the product goes to zero.
