Limit of the sequence $\frac{\sin{1}+\sin{2}+...+\sin{n}}{n}$

Question

As the title suggests we need to find: $$\lim_{n\to \infty}{\frac{\sin{1}+\sin{2}+...+\sin{n}}{n}}$$ What I did:

$$\sum_{k=1}^{n}{\sin{k}}=\operatorname{Im}\big(\sum_{k=0}^{n}{e^{ik}}\big)=\operatorname{Im}\big(\frac{1-e^{i(n+1)}}{1-e^i}\big)=\operatorname{Im}\big(\frac{-2i\sin\big(\frac{n+1}{2}\big)}{-2i\sin{\frac{1}{2}}}e^{i\frac{n}{2}}\big)$$ $$=\frac{\sin\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2}\right)}{\sin{\frac{1}{2}}}$$.

$$\left|\frac{1}{n}\sum_{k=1}^{n}{\sin{k}}\right|=\frac{1}{n}\left|{\frac{\sin\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2}\right)}{\sin{\frac{1}{2}}}}\right|\le \frac{1}{n\left|\sin{\frac{1}{2}}\right|}\longrightarrow_{\infty} 0$$

By the way I don’t like this method using lots of tools such as complex numbers, geometric series... Did someone has another way to solve, specially using integral estimate.

$\textbf{Edit}$ I have found an integral estimate.

By Abel’s Lemma :

$$\sum_{n\le x}{\sin{n}}=\lfloor{x}\rfloor\sin{x}-\int_{1}^{x}{\lfloor{t}\rfloor\cos{t}dt}$$ $$=x\sin{x}-\{x\}\sin{x}-\int_{1}^{x}{\big(t\cos{t}-\{t\}\cos{t}\big)dt}$$ $$=x\sin{x}-\{x\}\sin{x}-\int_{1}^{x}{t\cos{t}dt}+\int_{1}^{x}{\{t\}\cos{t}dt}$$ $$= x\sin{x}-\{x\}\sin{x}-\big\{x\sin{x}-\sin{1}-\sin{x}+\sin 1\big\} +\int_{1}^{x}{\{t\}\cos{t}dt}$$ $$=\underbrace{\big(1-\{x\}\big)\sin{x}+\int_{1}^{x}{\{t\}\cos tdt}}_{\text{all this is O(1)}}$$

For $x=m$ we will have: $$\sum_{n=1}^{m}{\sin{n}}=\sin{m}+\int_{1}^{m}{\{t\}\cos{t}dt}=O(1)$$

$$\therefore \frac{1}{m}\sum_{n=1}^{m}{\sin{n}}=O\left(\frac{1}{m}\right)$$ $\textbf{rEdit:}$

We can deduce from above: $$\sum_{k=1}^{n}{\sin{k}}=\int_{1}^{n+1}{\{t\}\cos{t}}dt \tag{•}$$

Is $(•)$ a known relation?

Possible duplicate of How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? — Micah, Apr 09 '19 at 15:43
The accepted answer to the duplicate target does not involve complex numbers (the price is that you need more trig identities instead). — Micah, Apr 09 '19 at 15:44
How can you compute a limit involving trigonometric functions without trig? What does that even mean? — Micah, Apr 09 '19 at 15:56
I think you'll find that any such integral has some trigonometric facts lurking somewhere in its derivation... — Micah, Apr 09 '19 at 16:11
Also, I don't think the bounds you get from an integral estimate are likely to be good enough. We have $\sum_{k=1}^n \sin n \approx \int_{1/2}^{n+1/2} \sin x , dx$, but the standard error estimate is $O((b-a)^3/n^2)=O(n)$, so the most that gets you is that your sequence is bounded. In order to know that it converges to zero, you have to argue that those errors actually cancel out, which requires some argument tailored to the sine function. — Micah, Apr 09 '19 at 16:25

score 5 · Accepted Answer · 2019-04-09T18:20:20.757

5

Hint:

$$S_n:=\sum_{k=1}^n\sin(k)=\sum_{k=0}^{n-1}\sin(k+1)=\sin(1)\sum_{k=0}^{n-1}\cos (k)+\cos(1)\sum_{k=0}^{n-1}\sin(k) \\=\sin(1)(C_n-\cos(n)+1)+\cos(1)(S_n-\sin(n)),$$

and $$C_n:=\sum_{k=0}^{n-1}\cos(k+1)=\cos(1)\sum_{k=0}^{n-1}\cos(k)-\sin(1)\sum_{k=0}^{n-1}\sin(k) \\=\cos(1)(C_n-\cos(n)+1)-\sin(1)(S_n-\sin(n)).$$

This is a system of two equations in two unknowns $S_n,C_n$. Solve and take the limit of

$$\dfrac{S_n}n.$$

edited Apr 09 '19 at 18:20

answered Apr 09 '19 at 15:53

1

Thank you for this – DINEDINE Apr 09 '19 at 15:54

Limit of the sequence $\frac{\sin{1}+\sin{2}+...+\sin{n}}{n}$

1 Answers1