Order of Möbius Function
Theorem
Let $\mu$ denote the Möbius function .
Then:
- $\displaystyle \sum_{n \mathop \le N} \map \mu n = \map o N$
where $o$ denotes little-o notation.
Proof
Let $\map \Re z$ 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 are Even Negative Integers, the Riemann zeta function has no zeroes in $\map \Re z > 1$.
Thus the reciprocal of the Riemann zeta function is analytic in $\map \Re z > 1$.
By Reciprocal of Riemann Zeta Function, this means that $\displaystyle \sum_{n \mathop = 1}^\infty \map \mu n n^{-z}$ converges to an analytic function in $\map \Re z > 1$.
By taking $a_n = \mu \left({n}\right)$ in Ingham's Theorem on Convergent Dirichlet Series:
- $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$ converges for $\map \Re z \ge 1$.
Taking $z = 1$, we are given a convergent sum:
- $\displaystyle \sum_{n \mathop = 1}^\infty \frac {\map \mu n} n$
Clearly:
- $\displaystyle \sum_{n \mathop = 1}^N \frac {\map \mu n} n \ge \sum_{n \mathop = 1}^N \frac {\map \mu n} N$
but:
- $\displaystyle \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} n = \lim_{z \mathop \to 1} \frac 1 {\map \zeta z}$
Since Harmonic Series is Divergent, $\dfrac 1 {\map \zeta z}$ goes to $0$ by Reciprocal of Null Sequence.
Hence also:
- $\displaystyle \lim_{N \mathop \to \infty} \sum_{n \mathop = 1}^N \frac {\map \mu n} N = 0$
$\blacksquare$