Order of Möbius Function

Theorem
If $\mu$ is the Moebius function and $o$ is little "o", then


 * $\displaystyle \sum_{n \mathop \le N} \mu(n) = o(N)$

Proof
Let $\Re \left({z}\right)$ be the real part of a complex variable $z$.

Since the Riemann zeta function is analytic and zero-free in $\Re \left({z}\right) > 1$ the inverse of the Zeta function, we know the inverse of the Zeta function is analytic in $\Re \left({z}\right) > 1$.

By the inverse of the Zeta function, this means that $\displaystyle \sum_{n=1}^\infty \mu(n) n^{-z}$ converges to an analytic function in $\Re \left({z}\right) > 1$.

By taking $a_n = \mu(n)$ in Ingham's theorem on convergent Dirichlet series, we see that:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mu(n)}{n^z}$

converges for $\Re \left({z}\right) \ge 1$.

Taking $z=1$, we are given a convergent sum:


 * $\displaystyle \sum_{n \mathop = 1}^\infty \frac{\mu(n)} n$

Clearly:


 * $\displaystyle \sum_{n \mathop = 1}^N \frac{\mu(n)} n \ge \sum_{n \mathop = 1}^N \frac{\mu(n)} N$

but


 * $\displaystyle \lim_{N \to \infty} \sum_{n \mathop = 1}^N \frac{\mu(n)} n = \lim_{z \to 1} \frac{1}{\zeta(z)}$

Since the harmonic series goes to infinity, $1/\zeta(z)$ goes to $0$ by Reciprocal of Null Sequence, and hence


 * $\displaystyle \lim_{N \to \infty} \sum_{n \mathop = 1}^N \frac{\mu(n)} N = 0 $

as well.