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$$\sum_{n>0} \frac{\mu(n)}{n^s}$$

Sum from 1 to infinity of The Möbius function$/n^s$, i.e., Möbius function/Riemman-zeta function?

Sorry, I forgot to mention that the way that I am suppose to tackle this question is with the information that: L(s,f[convolution]g) = L(s,f)*L(s,g).

The problem was that I couldn't use the mobius inversion theorem or anything to try and find the mobius function in terms of a convolution? Any help? thanks.

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  • $\begingroup$ Just a minor point: what you have written is not "Möbius/Zeta," although it sort of looks like it. $\endgroup$ – angryavian Jan 11 '14 at 18:32
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Hints for you to complete and prove:

Put $\;\mathcal P:= \{p\in\Bbb N\;;\;p\;\text{is a prime}\}\;$ :

$$\text{Re}(s)>1\;:\;\;\zeta(s)=\prod_{p\in\mathcal P}\left(1-\frac1{p^s}\right)^{-1}\implies$$

$$\frac1{\zeta(s)}=\prod_{p\in\mathcal P}\left(1-\frac1{p^s}\right)=1-\sum_{p\in\mathcal P}\frac1{p^s}+\sum_{p,q\in\mathcal P\,,\,p\neq q}\frac1{p^sq^s}-\sum_{p_1,p_2,p_3\in\mathcal P\,,\,i\neq j\implies p_i\neq p_j}\frac1{p_1^sp_2^sp_3^s}+\ldots$$

$$=\sum_{n=1}^\infty\frac{\mu(n)}{n^s}$$

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