Let $a_n$ denote the $n$-th composite squarefree number.
Then $$ \sum_n\frac{1}{a_n^s} = \frac{\zeta(s)}{\zeta(2s)}-1-P(s) $$ for $s>1$ (in fact, for $Re(s)>1$)
source and details:
https://en.wikipedia.org/wiki/Prime_zeta_function
http://mathworld.wolfram.com/PrimeZetaFunction.html
Question: what is the value of $\sum\frac{1}{a_n}$ ?
On
The sum diverges at $s = 1$. This is visible from the exporession you wrote down from the pole of the zeta function at $s = 1$.
You can heuristic it by thinking of the fact that about $6/\pi^2$ of all natural numbers are square-free [in the asymptotic density sense], so this sum should diverge at about the same rate as $\sum \frac{1}{n}$.
On
A partial sum of terms with denoms $\le N$ is bounded below by the sum-over-semiprimes
$$\sum_{p<q\le \sqrt{N}}\frac{1}{pq}=\frac{1}{2}\left( \big(\sum_{p\le\sqrt{N}}1/p\big)^2-\sum_{p\le\sqrt{N}}1/p^2 \right), $$
which diverges since $\sum_{p\le\sqrt{N}}1/p$ diverges and $\sum_{p\le\sqrt{N}}1/p^2$ converges.
It occurs to me Marco's answer eclipses this one in terms of simplicity. However, it is generally useful to know that $(\sum_{i\in I} a_i)^2=(\sum_{i\in I}a_i^2)+\sum_{i,j\in I} a_i a_j$. And perhaps this argument can be taken further to get an asymptotic expansion of the partial sum as a function of $N$.
Since $$\sum_{n\geq1}\frac{1}{a_{n}^{s}}=\frac{\zeta\left(s\right)}{\zeta\left(2s\right)}-1-P\left(s\right) $$ we have $$\sum_{n\geq1}\frac{1}{a_{n}}=\lim_{s\rightarrow1}\left(\frac{\zeta\left(s\right)}{\zeta\left(2s\right)}-1-P\left(s\right)\right) $$ and now since $\zeta\left(s\right)\sim\frac{1}{s-1}$ and $P\left(s\right)\sim\log\left(\frac{1}{s-1}\right)$ when $s\rightarrow 1$ we can observe that the series diverges.
The most rapid way to prove the divergence is observing that $$\sum_{n\geq1}\frac{1}{a_{n}}>\sum_{p>2}\frac{1}{2p} $$ since obviously every number in the form $2p,\, p>2$ is a composite squarefree number and since the series of the reciprocal of primes diverges we can conclude.