Proving $\sum\limits_{n=1}^\infty\frac{|\mu(n)|}{n^s}=\frac{\zeta(s)}{\zeta(2s)}.$

Question

I am trying to show that $$D(|\mu|,s)=\sum_{n=1}^\infty\frac{|\mu(n)|}{n^s}=\frac{\zeta(s)}{\zeta(2s)}.$$ By a previous exercise I know that $D(\lambda,s)=\zeta(2s)/\zeta(s)$, where $\lambda$ is Liouville's function $\lambda(n)=(-1)^{\Omega(n)}$. But I don't see how to use this fact (I am not sure I should, but it looks like it would be useful).

Answer 1

The coefficient of the Dirichlet series is multiplicative, so you only need to compare the Euler factors.

Answer 2

This is how I convinced myself of this fact. Here's an argument which utilizes the Euler product. So I am going to take as a given:

$$\prod _{p} 1-p^{-s}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s }=\frac{1}{\zeta(s)}, \hspace{1 cm} \prod _{p} 1+p^{-s}=\sum_{n=1}^\infty \frac{|\mu(n)|}{n^s } $$

Then $$\frac{\zeta(s)}{\zeta(2s)}= \frac{\prod_p 1-p^{-2s}}{\prod_p 1-p^{-s}}=\prod_{p} \frac{1-p^{-2s}}{1-p^{-s}}=\prod_p \frac{(1-p^{-s})(1+p^{-s})}{1-p^{-s}}=\prod_{p}1+p^{-s}=\sum_{n=1}^\infty \frac{|\mu(n)|}{n^s}$$