TITLE: Square-free zeta function zeros
QUESTION [6 upvotes]: It is a well known fact that the geometric series 
$$1+x+x^2+x^3+\ldots$$
has the following form
$$\frac{1}{1-x}$$
Another possible representation  is
$$\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$
This comes from  the identity
$$1+x+x^2+x^3+\ldots+x^{2^{k}}=\frac{1-x^{2^{k}+1}}{1-x}$$
now taking the numerator of the rhs we have
$$1-x^{2^{k}+1}=\left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)=\left(1-x^{2^{k}-1}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$
proceeding this way we eventually get
$$\left(1-x\right)\left(1+x\right)\left(1+x^{2}\right)\ldots\left(1+x^{2^{k}-2}\right)\left(1+x^{2^{k}-1}\right)\left(1+x^{2^{k}}\right)$$
Taking the limit for the geometric series 
$$\sum_{k=0}^{\infty}x^{k}=\prod_{k=0}^{\infty}\left(1+x^{2^{k}}\right)$$
Now taking the zeta function
$$\zeta(z)=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}+\frac{1}{p^{2z}}+\frac{1}{p^{3z}}+\ldots\right)$$
we can express it as 
$$\zeta(z)=\prod_{k=0}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$
Now considere  for 
$$G(z)=\prod_{k=1}^{\infty}\;\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z\;2^{k}}}\right)$$
note that now $k\geq 1$ and that $G(z)$ converges absolutely for $z>\frac{1}{2}$
Can we say that, after analytic continuation, that
$$H(z)=\sum_{k=0}^{^\infty}\frac{|\mu(k)|}{k^{z}}=\prod_{p\in\mathbb{P}}\left(1+\frac{1}{p^{z}}\right)$$ 
has exactly the same zeros as $\zeta(z)$?

REPLY [10 votes]: Hint: $$ \sum \frac{|\mu(k)|}{k^z} = \frac{\zeta(z)}{\zeta(2z)} $$ for $\Re(z) > 1$
Check out http://en.wikipedia.org/wiki/Dirichlet_series 
It might help you out.