Is the sum of the reciprocals of the squarefree numbers divergent or convergent? I just come across this question by trying to analyze the pseudoinverse of some infinite matrix (the matrix T as interpreted in my answer to this MSE-question), where this series occurs from some dotproducts. I think it can also be expressed in terms of the Moebius-function:
$$\sum_{k=1}^\infty \left| {\operatorname{moebius}(k) \over k} \right|  $$ 
Remembering, that the sum of the reciprocals of the primes is divergent I think I should assume divergence here as well, but I'm not sure.         


*

*Q1: Is this sum convergent (and if, what is its value)?            

*Q2: if it is divergent, is there possibly some finite value related, like for instance the Euler/Mascheroni-$\gamma$ for the harmonic series?

The asymtotic growth of the sum of the reciprocals of squarefree numbers $n\le x$ is $\frac{6 e^\gamma}{\pi^2} \log x$, because we have
$$
\prod\limits_{p≤x}\left(1 + \frac{1}{p} \right) = \frac{\prod\limits_{p≤x}\left(1 - \frac{1}{p^2} \right)}{\prod\limits_{p≤x}\left(1 - \frac{1}{p} \right)} \sim \frac{\frac{1}{\zeta(2)}}{\frac{1}{e^\gamma \log x}} =  \frac{6 e^\gamma}{\pi^2} \log x.
$$
In particular we have
$$
\sum_{n {\rm \; squarefree}}\frac{1}{n}=\prod_{p}\left(1 + \frac{1}{p} \right) =\infty.
$$
