Order of Möbius Function

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

Proof
By taking $$a_n = \mu(n) \ $$ in Inghams theorem on convergent Dirichlet series, we see that

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

converges for $$\Re \left({z}\right) \geq 1 \ $$, where $$\Re \left({z}\right)$$ is the real part of $$z \ $$. Taking $$z=1 \ $$, we are given a convergent sum

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