How prove or disprove this limit $\lim_{n\to\infty}\frac{\sum_{k=1}^{n}|\sin{(x+k)}|}{n}=\dfrac{1}{\pi}\int_{0}^{\pi}|\sin{x}|dx$

Question

prove or disprove $$\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=1}^{n}|\sin{(x+k)}|}{n}=\dfrac{1}{\pi}\int_{0}^{\pi}|\sin{x}|dx$$

if this is right,and How prove it? Thank you

My idea maybe this can use Uniform distribution? and I don't prove it.

Answer 1

Claim: If the continuous function $f(x)$ is periodic with period $T$ which is irrational, then

$$ \lim_{n\rightarrow \infty} \frac{ \sum_{k=1}^{n} f(t + k)}{n} = \frac{1}{T} \int_0^T f(x) \, dx$$

Hint: The RHS calculates the average value of $f(x)$ (over 1 period). Show that the LHS is equivalent to (approximating) the average value of $f(x)$.

Hint: This is an application of the pigeonhole argument which shows that for any $ \epsilon > 0$, there exists an integer $n$ such that $n T$ is within $ \epsilon$ of an integer.

Answer 2

Lemma 1. Let $\theta$ be a real number which is not a rational multiple of $\pi$. For $m\in\mathbb{Z}$, we define $$I_n(m)=\frac{1}{n}\sum_{k=1}^ne^{ikm\theta}$$ Then, $$\lim_{n\to\infty}I_n(m)=\left\{\matrix{0&\hbox{if}&m\ne0\cr1&\hbox{if}&m=0}\right. $$

Proof. Indeed, for $m\ne0$ we have $$\left\vert I_n(m)\right\vert=\frac{1}{n}\left\vert\frac{e^{im(n+1)\theta}-e^{im\theta}}{e^{im\theta}-1}\right\vert\leq\frac{1}{n\vert \sin(m\theta/2)\vert} $$ and the conclusion follows since $\theta\notin\pi\mathbb{Q}$.$\qquad\square$

Lemma 2. Let $\theta$ be a real number with $\theta\notin\pi\mathbb{Q}$. Let $P$ be a trigonometric polynomial, $P(t)=\sum_{m=-r}^r c_me^{imt}$, and define $$J_n(P,t)=\frac{1}{n}\sum_{k=1}^nP(t+k\theta)$$ Then, $$\lim_{n\to\infty}J_n(P,t)=c_0=\frac{1}{2\pi}\int_0^{2\pi}P(x)dx $$

Proof. In fact $$ J_n(P,t)=\sum_{m=-r}^{r}c_me^{imt}\left(\frac{1}{n}\sum_{k=1}^ne^{imk\theta}\right) =\sum_{m=-r}^{r}c_me^{imt}I_n(m) $$ and the conclusion follows by letting $n$ tend to infinity. $\qquad\square$

Proposition 3. Let $\theta$ be a real number with $\theta\notin\pi\mathbb{Q}$. Let $f$ be a continuous $2\pi$-periodic function, and define $$J_n(f,t)=\frac{1}{n}\sum_{k=1}^nf(t+k\theta)$$ Then, $$\lim_{n\to\infty}J_n(f,t)= \frac{1}{2\pi}\int_0^{2\pi}f(x)dx $$

Proof. Indeed, for $\epsilon>0$ there is a trigonometric polynomial $P_\epsilon$ such that $$ \Vert f-P_\epsilon\Vert_\infty=\sup_{\mathbb{R}}|f(t)-P_\epsilon(t)|<\frac{\epsilon}{3} $$ It follows that $$ \vert J_n(f,t)-J_n(P_\epsilon,t)\vert\leq\Vert f-P_\epsilon\Vert_\infty<\frac{\epsilon}{3}\tag{1} $$ and also $$ \left\vert \frac{1}{2\pi}\int_0^{2\pi}f(x)dx-\frac{1}{2\pi}\int_0^{2\pi}P_\epsilon(x)dx\right\vert\leq\Vert f-P_\epsilon\Vert_\infty<\frac{\epsilon}{3}\tag{2} $$ Finally, using Lemma 2. there is $N_\epsilon$ such that for $n>N_\epsilon$ we have $$ \left\vert J_n(P_\epsilon,t)-\frac{1}{2\pi}\int_0^{2\pi}P_\epsilon(x)dx\right\vert<\frac{\epsilon}{3}\tag{3} $$ Combining $(1)$, $(2)$ and $(3)$ we obtain $$ \forall\,n>N_\epsilon,\quad\left\vert J_n(f_\epsilon,t)-\frac{1}{2\pi}\int_0^{2\pi}f(x)dx\right\vert<\epsilon. $$ and the Proposition follows.$\qquad\square$

Application. Taking $f(x)=|\sin(x)|$, and $\theta=1$ we see that $$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n|\sin(t+k)|= \frac{1}{2\pi}\int_0^{2\pi}|\sin(x)|dx.$$