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Prove that for Re(s)>1

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

Where $\mu(n)$ is the Möbius function defined by:

$\mu(n)=1, \mbox{if }n=1$

$\mu(n)=(-1)^k, \mbox{if }n=p_1,p_2,...,p_k$ and $p_j$ are distinct primes

$\mu(n)=0, otherwise$

Hint: use the Euler product formula.

I started trying to use the Euler product formula for $\zeta(s)$ but no succes. How do I get this result?

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  • $\begingroup$ What actually does the Euler product give for $1/\zeta(s)$? $\endgroup$ – Lord Shark the Unknown Apr 28 '18 at 13:21
  • $\begingroup$ I used the product formula for $\zeta(s)$, question edited $\endgroup$ – tererecomchimarrao Apr 28 '18 at 13:24
  • $\begingroup$ Is a comparison of the Euler product that you can need to calculate, and the Euler product corresponding to $\zeta(s)$. You need to see in you book how to calculate the corresponding Euler product (for the completely multiplicative function $\mu(n)$). $\endgroup$ – user243301 Apr 28 '18 at 13:31
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Euler Product$$\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}$$

$$\frac{1}{\zeta(s)}=\prod_{p}(1-p^{-s})=(1-p_1^{-s})(1-p_2^{-s})(1-p_3^{-s})...$$

Due to the fundamental theorem of arithmetic, this product can be re-written as a sum over the integers. By foiling out the primes, each term in the sum can be seen to be made up of a product of different combinations of primes, and the sign of the term is based on the amount of prime factors it contains.

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

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