Is $a_n=\frac{1}{n}\sum_{k=1}^n\frac{\varphi(k)}{k}$ convergent? Let $(a_n)_{n\in\mathbb{N}}$ be defined as $a_n=\frac{1}{n}\sum_{k=1}^n\frac{\varphi(k)}{k}$ where $\varphi$ is the euler totient function. Is $(a_n)$ convergent. If so, what is its limit?
I have checked it numerically; it seems to converge to the value
$$
a\approx 0.6079384135652464404586775568799731914204890312331725
$$
However, I cannot think of a way to prove it.
 A: Here you can find that the value you're looking for is $ \frac{6}{\pi^2} $
A: Start from $$\sum_{d|n} \varphi(d) = n$$
which yields $$L_1(s) = \sum_{n\ge 1} \frac{\varphi(n)}{n^s}
= \frac{\zeta(s-1)}{\zeta(s)}
\quad\text{and}\quad
L_2(s) = \sum_{n\ge 1} \frac{\varphi(n)/n}{n^s}
= \frac{\zeta(s)}{\zeta(s+1)}.$$
We have at $s=1$ being a simple pole
$$\mathrm{Res}(L_2(s); s=1) = \frac{6}{\pi^2}$$
and therefore by the 
Wiener-Ikehara theorem
we obtain $$\frac{1}{n} \sum_{k=1}^n \frac{\varphi(k)}{k}
\sim \frac{1}{n} \frac{6}{\pi^2} n
= \frac{6}{\pi^2}.$$
Remark. This is a powerful theorem and hence we may or may not be permitted to apply it to this problem.
A: See a result from this paper or the sum in the wiki article.  The sum may be approximated:
$$\sum_{k=1}^n \frac{\varphi(k)}{k} = \frac{6}{\pi^2}n + \mathcal{O}\left((\log n)^{2/3}(\log \log n)^{4/3}\right)$$
Since, as $n\to \infty$, $n>(\log n)^{2/3}(\log \log n)^{4/3}$, the big-O term drops out.  Thus, we're left with the limit being $\frac{6}{\pi^2}$.
