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

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.

• I have tried,and I can't true this right,and this problem is find limit $\lim_{n\to\infty}\dfrac{\sum_{k=1}^{n}|\sin{(x+k)}}{n}$! – math110 Jun 4 '14 at 3:16
• I believe you need a factor of $\frac{1}{\pi}$ on the RHS. – Calvin Lin Jun 4 '14 at 3:24

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 $$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n|\sin(t+k)|= \frac{1}{2\pi}\int_0^{2\pi}|\sin(x)|dx.$$

• It's Nice! Thank you very much!+1 – math110 Jun 4 '14 at 11:03
• Now this is an answer. =) +1 – Pedro Tamaroff Jun 4 '14 at 21:59
• It's curious the result is $x$-independent. Is that a kind of Ergodic Theorem ?. – Felix Marin Aug 9 '14 at 21:57

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.

• Why the down vote? – Calvin Lin Jun 4 '14 at 3:17
• Hello,I think you hint is not usefull? because I have think it this hint,and I can't prove it? can you post your full solution? – math110 Jun 4 '14 at 3:18
• You'd have to prove the LHS is independent of the choice of $t$, I guess. – Pedro Tamaroff Jun 4 '14 at 3:27
• @PedroTamaroff Yes, that is true, and that's why I intentionally chose the variable $t$ instead of $x$. – Calvin Lin Jun 4 '14 at 3:32
• @math110 I'm not hard up for up votes, nor points. Thank you very much. – Calvin Lin Jun 4 '14 at 4:04