For a number theory problem, I'm trying to prove that the asymptotic probability of numbers being relative prime equals $\frac{6}{\pi^2}$. Unfortunately, I'm a bit stuck at the following equality.
$$\frac{n^2}{2}\sum_{d=1}^n \frac{\mu(d)}{d^2}-\frac{n}{2}\sum_{d=1}^n\frac{\mu(d)}{d} \leq \sum_{k=1}^n \phi(k) \leq \frac{n^2}{2}\sum_{d=1}^n \frac{\mu(d)}{d^2}+\frac{n}{2}\sum_{d=1}^n\frac{\mu(d)}{d} $$
Where $\mu(d)$ is the Möbius arithmetic function and $\phi(k)$ is the Euler $\phi$ function.
Now I know that $$|\sum_{d=1}^n\frac{\mu(d)}{d}| \leq \sum_{d=1}^n\frac{1}{d} = 1+\sum_{d=2}^n\frac{1}{d} \leq 1+\log{n},$$ However, what I don't quite get is why it then should follow that $$\lim_{n\to \infty} \frac{\sum_{k=2}^n\phi(k)}{n(n-1)} = \frac{1}{2}\sum_{d=1}^{\infty} \frac{\mu(d)}{d^2}.$$
Any help on why this is the case?