TITLE: Dirichlet series associated to squared Möbius
QUESTION [6 upvotes]: 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 $F/\mathbf{Q}$?

REPLY [3 votes]: 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 
\[\sum_{n \leq x} \frac{\mu(n)^2}{n} = A(x) + \frac{1}{\zeta(2)} \sum_{n \leq x} \frac{1}{n} = \frac{\log x+\gamma}{\zeta(2)} + F(1)+o(1).\]