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

esege
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    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. – Cardboard Box Apr 08 '14 at 20:44
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    Or find Euler product formula for that Dirichlet series. – Cortizol Apr 08 '14 at 20:51
  • @CardboardBox I know it's been 6 years but hopefully you're still active haha. How can we apply Mobius inversion here since the sum is from 1 to infinity and not over the divisors of n? I'm trying to prove this without the Euler product formula. – BalsamicVinegar Mar 29 '20 at 07:56
  • @BalsamicVinegar If you multiply two Dirichlet series $\sum a_n n^{-s}$ and $\sum b_n n^{-s}$, note that the result is $\sum c_n n^{-s}$, where $c_n = \sum_{d \mid n} a_d b_{n/d}$. – Cardboard Box Apr 02 '20 at 12:17

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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.

glebovg
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