Is $a_n=\frac{1}{n}\sum_{k=1}^n\frac{\varphi(k)}{k}$ convergent?

Question

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.

Answer 1

Here you can find that the value you're looking for is $ \frac{6}{\pi^2} $

Answer 2

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 $\frac{6}{\pi^2}$.

Answer 3

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.