# How to prove $\sum_{n=1}^{\infty}\frac{\mu (n)}{n^{s}}=\frac{1}{\zeta (s)}$?

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

• 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. – Dane Apr 8 '14 at 20:44
• Or find Euler product formula for that Dirichlet series. – Cortizol Apr 8 '14 at 20:51

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.