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.


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.
