Let $Q(x)$ denote the number of squarefree positive integers $n$ with $1\le n\le x$. Then we may write $Q(x)=\frac{6x}{\pi^2}+R(x)$, where $R(x)=O(\sqrt{x})$ as $x\to \infty$ ($R(x)$ is in fact smaller than this. See: wikipedia: square-free integer). Now we apply Abel's summation formula, to find
\begin{align*}\sum_{\substack{n\le x \\ n \text{ squarefree}}}\frac{1}{n}=&\;\frac{Q(x)}{x}+\int_{1}^{x}\frac{Q(t)}{t^2}\text{d}t \\[-1mm] =&\; \frac{6}{\pi^2}+\frac{R(x)}{x}+\frac{6}{\pi^2}\int_{1}^{x}\frac{\text{d}t}{t}+\int_{1}^{x}\frac{R(t)}{t^2}\text{d}t \\[2mm] =&\; \frac{6\log x}{\pi^2}+\Big(\frac{6}{\pi^2}+\int_{1}^{\infty}\frac{R(t)}{t^2}\text{d}t\Big)+\Big(\frac{R(x)}{x}-\int_{x}^{\infty}\frac{R(t)}{t^2}\text{d}t\Big).\end{align*}
The first bracketed term converges as $x\to \infty$ (to $C_{\text{sf}}$, say), while the second bracketed term vanishes as $x\to \infty$ (the strength of the bound may depend on if the Riemann hypothesis is assumed or not).
The constant $C_{\text{sf}}$ can be expressed more transparently as follows:
First note the identity
$$(\ast) \hspace{1cm} \sum_{n\le x}\sum_{d|n}f(d)=\sum_{d\le x}f(d)\Big\lfloor \frac{x}{d}\Big\rfloor=x\sum_{d\le x}\frac{f(d)}{d}-\sum_{d\le x}f(d)\Big\{\frac{x}{d}\Big\}.$$
Second, note that we can write
$$Q(x)=\sum_{n\le x}\sum_{d|n}f(d),$$
where
$$f(d)=\begin{cases}\mu(\sqrt{d}) & \text{if } d \text{ is a perfect square,} \\ 0 & \text{otherwise.} \end{cases}$$
Using this in the identity $(\ast)$, we obtain
\begin{align*}Q(x)=&\; x\sum_{d^2\le x}\frac{\mu(d)}{d^2}-\sum_{d^2\le x}\mu(d)\Big\{\frac{x}{d^2}\Big\} \\
=&\; \frac{6x}{\pi^2} \hspace{2mm}\underbrace{-x\sum_{d^2>x}\frac{\mu(d)}{d^2}-\sum_{d^2\le x}\mu(d)\Big\{\frac{x}{d^2}\Big\}}_{R(x)}. \end{align*}
In summary, the squarefree harmonic series diverges to $+\infty$, is asymptotic to $ \frac{6\log x}{\pi^2}$, and has an associated constant $C_{\text{sf}}$, where
\begin{align*}C_{\text{sf}}=&\;\lim_{x\to \infty}\Big(-\frac{6\log x}{\pi^2}+\sum_{\substack{n\le x \\ n \text{ squarefree}}}\frac{1}{n}\Big)=\frac{6}{\pi^2}+\int_{1}^{\infty}\frac{R(t)}{t^2}\text{d}t \\[1.5mm]
=&\; \frac{6}{\pi^2}-\int_{1}^{\infty}\frac{1}{t^2}\bigg(\sum_{d^2>t}\frac{\mu(d)}{d^2}+\sum_{d^2\le t}\mu(d)\Big\{\frac{t}{d^2}\Big\}\bigg) \text{d}t.\end{align*}
Some parts of the calculation above may be found with a more detailed explanation in Montgomery & Vaughan's Multiplicative Number Theory I: Classical Theory (Chapter 2.1).
