$$\sum_{n=1}^{\infty} \frac{λ(n)}{n^s}=\frac{ζ(2s)}{ζ(s)}$$
Let $λ(n) = (−1)^k$, where $k$ is the number of prime factors of $n$, counting multiplicities. (Liouville function) for $Re(s)>1$, where $ζ(s)$ is the Riemann zeta function.
I am trying to prove this and I don't know where to start.
Since $\lambda$ Is multiplicative:
$$\begin{align}\sum_{n=1}^\infty \frac{\lambda(n)}{n^s}&=\prod_p\left(\sum_{k=0}^{\infty} \frac{\lambda(p^k)}{p^{ks}}\right)\\ &=\prod_p \left(\sum_{k=0}^{\infty} \frac{(-1)^k}{p^{ks}}\right)\\ &=\prod_p \left(\frac1{1+\frac 1{p^s}}\right)\\ &=\prod_p \frac{1-\frac1{p^s}}{1-\frac1{p^{2s}}}\\ &=\prod_p \dfrac{\dfrac{1}{1-\frac1{p^{2s}}}}{\dfrac1{1-\frac1{p^s}}}\\ &=\frac{\zeta(2s)}{\zeta(s)} \end{align}$$