2
$\begingroup$

How can we prove this equation? $$\sum_{n=1}^{\infty}\frac{\mu (n)}{n^{s}}=\frac{1}{\zeta (s)}$$

$\endgroup$
  • 3
    $\begingroup$ Are you familiar with Mobius inversion? If you are, then try multiplying the series for $\zeta(s)$ by $\sum \mu(n)/n^s$ and expanding. $\endgroup$ – Dane Apr 8 '14 at 20:44
  • 2
    $\begingroup$ Or find Euler product formula for that Dirichlet series. $\endgroup$ – Cortizol Apr 8 '14 at 20:51
2
$\begingroup$

Let $a(n)$ be a multiplicative number-theoretic function function. Then we have $$\sum_{n = 1}^\infty \frac{a(n)}{n^s} = \prod_{p \text{ prime}} \{1 + a(p)p^{-s} + a(p^2)p^{-2s} + \cdots\}, \quad \operatorname{Re}[s] \geq s_0,$$ which is known as the Euler product formula. The equality above is not difficult to verify. Suppose $a(n) = \mu(n)$. Then \begin{align*} \sum_{n = 1}\frac{\mu(n)}{n^s} &= \prod_{p \text{ prime}} \{1 + \mu(p)p^{-s} + \mu(p^2)p^{-2s} + \cdots\}\\ &= \prod_{p \text{ prime}} \{1 - p^{-s}\}, \end{align*} for it is obvious that $\mu(p) = -1$ and $\mu(p^s) = 0$ for $s = 2, 3, 4, \ldots$ whenever $p$ is prime. But $$\prod_{p \text{ prime}} \{1 - p^{-s}\} = \frac{1}{\zeta(s)},$$ so we are done.

$\endgroup$

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.