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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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    How can you say that G(z) converges for $z > \frac{1}{2}$ ?2011-03-03
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    @Roupa: because $w=2z,4z,8z,\ldots$.2011-03-03
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    what is $w$? And how do you say that the expression converges for $\Re(z) > \frac{1}{2}$? Can you please elaborate?2011-03-03
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    @roupam, all the powers are multiples of a power of 2.2011-03-04

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