Prove that for Re(s)>1


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?

  • $\begingroup$ 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

Euler Product$$\zeta(s)=\prod_{p}\frac{1}{1-p^{-s}}$$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.