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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 $\zeta(z)$?

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    @roupam, all the powers are multiples of a power of 2.2011-03-04

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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.

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    @Neves: You can factor $\sum \mu(n)^2 n^{-s}$ as the Euler product $\prod (1+p^{-s})$. To get the identity Roupam mentions, note that $1+x = (1-x^2)/(1-x)$ and then use the product of the original Riemann zeta function. Assuming RH, all zeros are therefore on $\zeta$'s critical line.2011-07-31