In the definition of Dirichlet generating function "for the square-free numbers " is: $$ \frac{\zeta(s)}{\zeta(2s)}=\sum_{n=1}^{\infty} \frac {|\mu(n)|}{n^s} $$
where $\mu$ is Moebius function .
Now suppose that: $$\sum_{n=1}^{\infty} \frac {|\mu(n)|}{n^s} =0 $$
then it is possible for $\mu(n)=0$ this implies $n$ is not a square free
and in the same time $\zeta(s)=0 $ according to the identity above .
My question here : How the definition of Dirichlet generating function Given
for square free number $n$ and in the same time no exception in the above
identity if RH were true ?
Thank you for any help
The identity:
$$ \text{Re}(s)>1,\qquad \frac{\zeta(s)}{\zeta(2s)}=\sum_{n\geq 1}\frac{\mu^2(n)}{n^s} $$ follows from Euler's product. Obviously $\mu^2(n)$ is a multiplicative function $\in\{0,1\}$, hence for any $s$ such that $\text{Re}(s)>1$ we have: $$\sum_{n\geq 1}\frac{\mu^2(n)}{n^s}=\prod_{p}\left(1+\frac{1}{p^s}\right)^{-1}=\frac{\prod_{p}\left(1-\frac{1}{p^s}\right)^{-1}}{\prod_{p}\left(1-\frac{1}{p^{2s}}\right)^{-1}}=\frac{\zeta(s)}{\zeta(2s)}.$$ Euler's product also gives that $\zeta(s)$ cannot vanish on $\text{Re}(s)>1$.