Dirichlet series associated to squared Möbius I would like to estimate the Dirichlet series of a multiplicative function. Consider the following:

$$\sum_{m \leqslant X} \frac{\mu^2(m)}{m^s}$$

When does it converges when $X$ grows? What is an equivalent for $s=1$?
More generally I have no method for doing so, obtaining an estimate of summatory functions or Dirichlet series knowing properties about the sum over divisors. Is there any suitable transform doing so?
If I generalize the settings a little, what can be said about the relation between such an estimate and the associated one on a more general number field? For instance, do we know something precise about:

$$\sum_{N\mathfrak{m} \leqslant X} \frac{\mu^2(\mathfrak{m})}{N\mathfrak{m}^s}$$

where $\mathfrak{m}$ runs along ideals of the integer ring of a number field $F/\mathbf{Q}$?
 A: Perron's inversion formula states that if $F(x) = \sum_{n \leq x} f(n)$ and $F(s) = \sum_{n = 1}^{\infty} f(n) n^{-s}$, then
\[F(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} F(s) x^s \, \frac{ds}{s}\]
for $\sigma$ sufficiently large and $x > 1$ not an integer. We take $f(n) = \sum_{N(\mathfrak{a}) = n} \mu^2(\mathfrak{a}) N(\mathfrak{a})^{-\alpha}$. Then via multiplicativity,
\[F(s) = \prod_{\mathfrak{p}} \sum_{k = 0}^{\infty} \frac{\mu^2(\mathfrak{p}^k)}{N(\mathfrak{p}^k)^{\alpha}} \frac{1}{k N(\mathfrak{p}^k)^s} = \prod_{\mathfrak{p}} \left(1 + \frac{1}{N(\mathfrak{p})^{\alpha + s}}\right) = \prod_{\mathfrak{p}} \frac{1 - N(\mathfrak{p})^{-2(\alpha + s)}}{1 - N(\mathfrak{p})^{-(\alpha + s)}} = \frac{\zeta_K(\alpha + s)}{\zeta_K(2(\alpha + s))},\]
where $\zeta_K(s)$ denotes the Dedekind zeta function of the number field $K$. So
\[\sum_{N(\mathfrak{a}) \leq x} \frac{\mu^2(\mathfrak{a})}{N(\mathfrak{a})^{\alpha}} = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \frac{\zeta_K(\alpha + s)}{\zeta_K(2(\alpha + s))} x^s \, \frac{ds}{s}\]
for $\sigma > \max\{1 - \alpha,0\}$. The integrand has a pole at $s = 1 - \alpha$, at $s = 0$, and at $s = \frac{\rho}{2} - \alpha$ for any zero $\rho$ of $\zeta_K(s)$. By Cauchy's residue theorem, we may move the line of integration to the left, picking up residues along the way. These give the main terms for the left-hand side.
Suppose first that $\alpha \neq 1$. Then the residue of the integrand as $s = 1 - \alpha$ is
\[\lim_{s \to 1} (s - 1) \zeta_K(s) \times \frac{x^{1 - \alpha}}{(1 - \alpha) \zeta_K(2)}.\]
When $K = \mathbb{Q}$, the limit is $1$ and $\zeta(2) = \pi^2/6$. In general, the limit is given by the analytic class number formula, and $\zeta_K(2)$ may not have such a nice closed form.
Now suppose that $\alpha = 1$. Then the integrand has a double pole at $s = 0$, so this residue is a little more complicated to calculate. We write the integrand in the form
\[\frac{\zeta_K(s + 1)}{\zeta_K(2s + 2)} \frac{e^{s \log x}}{s}\]
and expand this as a Laurent series around $s = 0$. The main term (after writing $e^{s \log x} = 1 + s \log x + O_x(s^2)$) turns out to be
\[\lim_{s \to 1} (s - 1) \zeta_K(s) \times \frac{\log x}{\zeta_K(2)}.\]
There is also a constant term that is much more complicated (at least when $K$ is not $\mathbb{Q}$).
A: We have that
\[\sum_{n=1}^\infty \frac{\mu(n)^2}{n^s} = \frac{\zeta(s)}{\zeta(2s)} = \sum_{n = 1}^{\infty} \frac{1}{n^s} \sum_{d^2 \mid n} \mu(d)\]
so that
\[\sum_{n \leq x} \mu(n)^2  = \sum_{n \leq x} \sum_{d^2 \mid n} \mu(d) = \sum_{d^2 \leq x} \mu(d) \left\lfloor \frac{x}{d^2} \right\rfloor = x \sum_{d^2 \leq x} \frac{\mu(d)}{d^2} + \sum_{d^2 \leq x} \mu(d) \left(\left\lfloor \frac{x}{d^2} \right\rfloor - \frac{x}{d^2}\right)\]
where
\[\sum_{d^2 \leq x} \frac{\mu(d)}{d^2} = \sum_{d = 1}^{\infty} \frac{\mu(d)}{d^2} - \sum_{d^2 > x} \frac{\mu(d)}{d^2} = \frac{1}{\zeta(2)} + O\left(\frac{1}{\sqrt{x}}\right)\]
and
\[\sum_{d^2 \leq x} \mu(d) \left(\left\lfloor \frac{x}{d^2} \right\rfloor - \frac{x}{d^2}\right) = O\left(\sqrt{x}\right),\]
i.e.
\[\sum_{n \leq x} \mu(n)^2  = \frac{x}{\zeta(2)} + O\left(\sqrt{x}\right).\]

Now let
\[B(x) = \sum_{n \leq x} \left(\mu(n)^2 - \frac{1}{\zeta(2)}\right) = O\left(\sqrt{x}\right), \qquad A(x) = \sum_{n \leq x} \frac{\mu(n)^2-\frac{1}{\zeta(2)}}{n}.\]
By summation by parts,
\[F(s) = \sum_{n=1}^\infty \frac{1}{n^s} \left(\mu(n)^2-\frac{1}{\zeta(2)}\right) = \lim_{x \to \infty} \left(\frac{B(x)}{x^s} - s \int_{1}^{x} \frac{B(t)}{t^s} \, \frac{dt}{t}\right)\]
converges for $Re(s) > 1/2$, so taking $s = 1$ shows that $F(1) =\lim_{x \to \infty} A(x)$ and $A(x) = F(1)+o(1)$, and so 
\[\sum_{n \leq x} \frac{\mu(n)^2}{n} = A(x) + \frac{1}{\zeta(2)} \sum_{n \leq x} \frac{1}{n} = \frac{\log x+\gamma}{\zeta(2)} + F(1)+o(1).\]
