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Evaluate the limit:

$$\lim_{n \rightarrow +\infty} \frac{1}{n^2} \sum_{k \leq n} \varphi(k)$$

where $\varphi$ is the Euler's function.

We use that \[\displaystyle\mathop{\sum}\limits_{k=1}^n\varphi(k)=\frac{3n^2}{\pi^2}+{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\]
Because \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)=0\quad (*)\] we have that
\begin{align*}
\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\mathop{\sum}\limits_{k=1}^n\varphi(k)&=\frac{3}{\pi^2}+\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\\
&=\frac{3}{\pi^2}+0\\
&=\frac{3}{\pi^2}\,.
\end{align*}


\((*)\) Left as an exercise.