Order of Möbius Function

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


 * $$\sum_{n \leq N} \mu(n) = o(N) \ $$

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

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 $$\sum_{n=1}^\infty \mu(n) n^{-z} \ $$ 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:


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

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

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


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

Clearly:


 * $$\sum_{n=1}^N \frac{\mu(n)}{n} \geq \sum_{n=1}^N \frac{\mu(n)}{N} \ $$

but


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

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


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

as well.