Euler Totient Series

We note that $\begin{aligned} \sum_{k=1}^n \varphi(k) & = \; \sum_{k=1}^n \sum_{d|k} \mu(d)\tfrac{k}{d} \; = \; \sum_{d=1}^n \mu(d) \sum_{m \le \frac{n}{d}}m \; = \; \sum_{d=1}^n \mu(d)\left[\frac{n^2}{2d^2} + \mathrm{O}\big(\tfrac{n}{d}\big)\right] \; = \; \tfrac12n^2 \sum_{d=1}^n \frac{\mu(d)}{d^2} + \mathrm{O}(n \ln n\big) \hspace{2cm} n \to \infty \end{aligned}$ while $\sum_{d=1}^n \frac{\mu(d)}{d^2} \; = \; \sum_{d=1}^\infty \frac{\mu(d)}{d^2} + \mathrm{O}\big(n^{-1}\big) \; = \; \zeta(2)^{-1} + \mathrm{O}\big(n^{-1}\big) \hspace{2cm} n \to \infty$ and hence $\sum_{k=1}^n \varphi(k) \; = \; \tfrac{3}{\pi^2}n^2 + \mathrm{O}(n \ln n\big) \hspace{2cm} n \to \infty$ making the desired limit equal to $\boxed{\tfrac{3}{\pi^2} = 0.3039635509}$ .