Order of Möbius Function

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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:

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

$\blacksquare$