Omega

Firstly we notice $\omega(n)$ is an additive function. It's easy enough to prove so I'll leave it as an exercise.

Since it is additive it implies $2^{\omega(n)}$ is a multiplicative function.

Let's make it general by considering:

$f(s)=\sum\limits_{n=1}^{\infty} \frac{2^{\omega(n)}}{n^s}= \prod_{p}\left[ 1+\frac{2^{\omega(p)}}{p^s}+\frac{2^{\omega(p^2)}}{p^{2s}}+...\right]$

Since $\omega(p^k)$ =1 we can simplify our expression by a geometric series to:

$f(s)=\prod_{p} \left[\frac{2}{1-\frac{1}{p^s}}-1\right]=\prod_{p} \frac{1+\frac{1}{p^s}}{1-\frac{1}{p^s}}=\prod_{p} \frac{1-\frac{1}{p^{2s}}}{\left(1-\frac{1}{p^s}\right)^2}=\frac{\zeta(s)^2}{\zeta(2s)}$

So in our case for $s=2$ we have $\large f(2)=\frac{\zeta(2)^2}{\zeta(4)}=\frac{\left(\frac{\pi^2}{6}\right)^2}{\frac{\pi^4}{90}}=\frac{5}{2}$ .

Use the observation that $(|\mu| * 1)(n) = 2^{\omega(n)}$ , the respective Dirichlet series for $|\mu|$ and $1$ ( $\dfrac{\zeta(s)}{\zeta(2s)}$ and $\zeta(s)$ respectively), and finally that $\displaystyle \left[ \sum_{m=1}^\infty \frac{f(m)}{m^s} \right] \left[ \sum_{n=1}^\infty \frac{g(n)}{n^s} \right] = \sum_{n=1}^\infty \frac{(f*g)(n)}{n^s}$ .