I want to prove that $$ \frac{1}{\zeta(s)}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}.$$

I know that the standard proof works with the Euler product formula $$\zeta(s)=\prod_{p \ \text{prime}} \frac{1}{1-p^{-s}}$$ but I am not really used to it (because I don't understand the proof for the Euler product formula). Now I want to know if there is another proof for the identity $\frac{1}{\zeta(s)}=\sum_{n=1}^\infty \frac{\mu(n)}{n^s}$. I thought there might be a method using Möbius inversion.


Let $\phi$ a multiplicative function and $\phi^*$ its convolutative inverse, that is $$ \sum\limits_{n|k} \phi(n) \phi^*(n/k) = \begin{cases} 1, & n=1, \\ 0, & else.\end{cases}$$

Assuming some mild growth condition on $\phi$ and $\phi'$, then we can define $$Z(\phi,s) = \sum\limits_{n=1}^\infty \phi(n) n^{-s}$$ and $ Z(\phi^*,s)$ analogous for suffienctly large $s \gg 0$ with absolute convergence guaranteed, then for all $s$ $$Z(\phi,s) Z(\phi^*,s) =1.$$

The idea of proof is $$ Z(\phi,s) Z(\phi^*,s) = \sum_{n,m} \frac{\phi(n) \phi^*(m)}{n^s m^s}$$ and then sum over $k=nm$ first (reorder the sum by absolute convergence). $$ \dots = \sum_{k=1}^\infty k^{-s} \sum\limits_{n|k} \phi(n) \phi^*(n/k) = 1.$$ You get the identity for $s \gg 0$ Then use uniqueness of analytic continuation to make sense for the remaining $s$.

  • $\begingroup$ I would like to have a formal, exact proof - not those dots $\ldots$. Because this is the similar method as with the euler product. $\endgroup$ – user140336 Apr 4 '14 at 14:09
  • 2
    $\begingroup$ Use induction... I am not here to do all of your homework;) $\endgroup$ – Marc Palm Apr 4 '14 at 14:38
  • $\begingroup$ You also need to proof convergence. $\endgroup$ – Marc Palm Apr 4 '14 at 14:46
  • $\begingroup$ I am not looking for a proof by induction... $\endgroup$ – user140336 Apr 4 '14 at 14:56
  • $\begingroup$ Okay, I got your point. $\endgroup$ – Marc Palm Apr 4 '14 at 18:09

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