limit of arithmetic mean of $|\sin n|$, $n\in\mathbb{N}$ In this posting, it is stated that

it is well known that $$ \lim_{n\rightarrow\infty}\frac{|\sin 1|+\ldots +|\sin n|}{n}=\frac{2}{\pi} $$ which can be obtained by the
uniform distribution.

I have tried to work that out but I can't get too far.
I tried instead to look at the problem as a Riemann sum
$$\frac{1}{n}\sum^n_{k=1}|\sin k|=\frac{1}{n}\sum^n_{k=1}\big|\sin \big(n\tfrac{k}{n}\big)\big|\approx \int^1_0|\sin nt|\,dt=\frac{1}{n}\int^n_0|\sin u|\,du$$
Tha last integral can be broken down in integrals over $(k\pi,(k+1)\pi]$, $k=1,\ldots \big[\tfrac{n}{\pi}\big]-1$, and a residual integral over $\Big[\big[\tfrac{n}{\pi}\big],n\Big]$. The residual is bounded by $\pi$ and so, when hit by $1/n$ it goes to $0$. The other integrals amount to $\big[\tfrac{n}{\pi}\big]2$ and so,
$$\frac{1}{n}\int^n_0|\sin u|\,du\xrightarrow{n\rightarrow\infty}\frac{2}{\pi}$$
This however does not answer fully the problem since the Riemann sum I propose is not for an integral over a fixed interval.
Can anybody give another suggestion?
 A: I am not aware thus far if there is a simple Calculus argument for this problem. Here is I present an argument that requires basic knowledge of Ergodic theory.

Consider the unit circle $\mathbb{S}^1=\{z\in\mathbb{C}:|z|=1\}$ equipped with the $\sigma$-algebra inherited as a subspace of $\mathbb{R}^2$ and the measure $\lambda_{\mathbb{S}^1}:=\frac{1}{2\pi}\lambda_1$, where $\lambda_1$ is the the arch-length measure. (Equivalently, this is the space  $[0,1]\mod 2\pi$ with the Borel $\sigma$-algebra and Lebesgue measure restricted to $[0,1]$.)
For any $\theta\in(0,1)$ define the rotation map
$$ 
\begin{align}
R_\theta&:\mathbb{S}^1\rightarrow\mathbb{S}^1\\
z&\mapsto z e^{2\pi i\theta}
\end{align}
$$
As you can easily see, $R_\theta$ is invariant with resect to $\lambda_{\mathbb{S}^1}$, that is $\lambda_{\mathbb{S}^1}\big(R^{-1}_{\theta}(A)\big)=\lambda_{\mathbb{S}^1}(A)$, that is rotating the circle by some angle $\theta$ does not change the length of an arch.

A little ergodic theory:
Theorem (Weyl): Suppose $R_\theta$ is an irrational rotation. Then, for every bounded Riemann integrable function $f:\mathbb{S}^1:\rightarrow\mathbb{C}^1$, and any $z\in\mathbb{S}^1$,
$$
\lim_\limits{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}f\circ R^k_\theta(z)=\int_{\mathbb{S}^1} f(w)\,\lambda_{\mathbb{S}^1}(dw)
$$
where $R^0_\theta(z)=z$ and $R^k_\theta(z)=R_\theta(R^{k-1}_\theta(z))$ for all $z\in\mathbb{S}^1$ and $k\geq1$.

Back to your problem.
Consider the rotation $R_\theta$ with $\theta=\frac{1}{2\pi}$. For $z=e^{ix}$,
$$
R^k_\theta(z)=ze^{2\pi i\theta k}=e^{i(x + k)}
$$
Consider the function $f:\mathbb{Z}\rightarrow\mathbb{R}$ given by $f(z)=|\operatorname{Im}(z)|$. Then
$$
f(R^k_\theta(z))=|\sin(x+k)|
$$
Since $f$ is continuous and thus, integrable, we can use Weyl's theorem to obtain that
$$
\lim_{n\rightarrow\infty}\frac{1}{n}\sum^{n-1}_{k=0}|\sin(x+k)| = \frac{1}{2\pi}\int^{2\pi}_0|\sin x|\,dx=\frac{2}{\pi}
$$
for all $x$. In your case, $x=1$.

Edit: Since Weyl's theorem can be proved with only tools from Calculus, I think that It would be worth while to add a short proof:
Proof of Weyl's theorem:
For any function $f$ on $\mathbb{S}^1$, let $S_nf(z)=\frac{1}{n}\sum^{n-1}_{j=1}f(R^j_\theta(z))$. Consider polynomials $f_k(z)=z^k$, $k\in\mathbb{\mathbb{Z}}$.
For $k=0$, $$S_nf_0(x)\equiv1=\frac{1}{2\pi}\int^{2\pi}_0 f_0
$$
For $|k|\geq1$,
$$S_nf_k(z)\equiv1=\frac{z^k}{n}\sum^{n-1}_{j=0}e^{i2\pi\theta k j}=
\frac{z^k}{n}\frac{1-e^{in 2\pi k\theta}}{1-e^{i2\pi k\theta}}\xrightarrow{n\rightarrow\infty}0=\int_{\mathbb{S}^1}f_k\,\lambda_{\mathbb{S}^1}=\frac{1}{2\pi}\int^{2\pi}_0 e^{ikx}\,dx
$$
since $e^{i2\pi k\theta}\neq0$ for all $k\in\mathbb{Z}$.
This means that the statement holds for all trigonometric polynomials. By the (complex) Stone-Weierstrass theorem, the result then holds for any $f\in\mathcal{C}(\mathbb{S}^1)$.
To extend the result to any  Riemann integrable function $f$, it is enough to assume that $f$ is real valued. Given $\varepsilon>0$, we can choose continuous functions $g_\varepsilon$ and $h_\varepsilon$ such that
$ g_\varepsilon < f\leq h_\varepsilon$,  and
$$ \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} -\varepsilon <\int_{\mathbb{S}^1} g_\varepsilon\, \lambda_{\mathbb{S}^1}\leq \int_{\mathbb{S}^1}h_\varepsilon \,\lambda_{\mathbb{S}^1}<  \int_{\mathbb{S}^1}f\,\lambda_{\mathbb{S}^1} +\varepsilon
$$
Then
$$
S_ng_\varepsilon-\int g_\varepsilon -\varepsilon \leq S_nf(z)-\int f \leq  S_nh_\varepsilon(z)-\int h_\varepsilon+\varepsilon
$$
whence we conclude that $$-\varepsilon\leq \liminf_{n\rightarrow\infty}S_nf(z)-\int f\leq\limsup_n S_nf-\int f\leq\varepsilon$$
for all $\varepsilon>0$ and $z\in\mathbb{S}^1$. This completes the proof.

Comment: Weyl's result is a little stronger than the plain Ergodic theorem since in the former there are no $\mu$-a.s. exceptional points where convergence fails (even though, were there any, they would form a set of $\lambda_{\mathbb{S}^1}$-measure zero).
A: Suggestion:
$$|\sin(k)|=|\sin(k\bmod\pi)|$$ and you can integrate over $[0,\pi)$.
