Order of Möbius Function

Theorem
Let $\mu$ denote the Möbius function.

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

where $o$ denotes little-o notation.

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

By Dirichlet Series is Analytic, the Riemann zeta function is analytic.

By Trivial Zeroes of Riemann Zeta Function, the Riemann zeta function has no zeroes in $\Re \left({z}\right) > 1$.

Thus the reciprocal of the Riemann zeta function is analytic in $\Re \left({z}\right) > 1$.

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

By taking $a_n = \mu \left({n}\right)$ in Ingham's Theorem on Convergent Dirichlet Series:


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

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

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


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

Clearly:


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

but:


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

Since Harmonic Series is Divergent, $\dfrac 1 {\zeta \left({z}\right)}$ goes to $0$ by Reciprocal of Null Sequence.

Hence also:


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