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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).

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The coefficient of the Dirichlet series is multiplicative, so you only need to compare the Euler factors. –  Sanchez Dec 8 '12 at 9:01
    
@Sanchez: Thank you. If you write it as an answer I will accept it. –  Carolus Dec 8 '12 at 9:50
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up vote 1 down vote accepted

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

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