\lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n \mu^2(k) = \frac{6}{\pi^2}.

\begin{aligned} \sum_{k = 1}^n \mu^2(k) &= \sum_{k = 1}^n \sum_{d^2 \mid k} \mu(d) \\ &= \sum_{d^2 \le n} \sum_{d^2 \mid k, k\le n} \mu(d) \\ &= \sum_{d^2 \le n} \mu(d) \sum_{ld^2 \le n} 1 \\ &= \sum_{d^2 \le n} \mu(d) \left(\frac{n}{d^2} + O(1)\right) \\ &= \sum_{d = 1}^{\left \lfloor \sqrt{n} \right \rfloor} \mu(d) \frac{n}{d^2} + O(1)\sum_{d = 1}^{\left \lfloor \sqrt{n} \right \rfloor} \mu(d) \\ &= \frac{n}{\zeta(2)} + O\left(\sqrt{n}\right) = \frac{6}{\pi^2}n + O \left(\sqrt{n}\right). \end{aligned}