Find $\lim_{n \to \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}$ [closed]

Another more general approach:

$$a_n\xrightarrow[n\to\infty]{} a\implies \frac{a_1+...+a_n}n\xrightarrow[n\to\infty]{}a$$

And since

$$\lim_{n\to\infty}\,n\,\sin\frac1n=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}=1\;\;\ldots\ldots$$

For small $|x|$ we have $\sin(x)\approx x$. Hence \[ i \cdot \sin\frac{1}{i} \approx 1 \] for big values of $i$. Hence \[ \frac{\sum_{i=1}^n i\cdot \sin\frac{1}{i} }{n}\approx \frac{n}{n}=1\]

Or we can use Stolz–Cesàro theorem to find that $$\lim_{n \rightarrow \infty}\frac{\sin 1+2\sin \frac{1}{2}+\cdots+n\sin \frac{1}{n}}{n}=\lim_{n\to \infty}\frac{(n+1)\sin {\frac{1}{n+1}}}{1}=1.$$

This is similar with @DonAntonio's solution.

Start by writing it in summation form. That is,

$$ \lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n} $$ Now, $\sin(1/i)<1/i$ for $i\in\mathbb{N}$. Furthermore, $\sin(1/i)>\frac1i-\frac1{6i^3}$ for $i\in\mathbb{N}$. Now, we squeeze the result by noting that $$ \lim_{n\to\infty} \sum_{i=1}^n \frac1n = 1 $$ and $$ \lim_{n\to\infty} \sum_{i=1}^n \frac{1-\frac1{6i^2}}n = \lim_{n\to\infty} \sum_{i=1}^n \frac1n-\frac1{6i^2n} = 1 $$ Therefore, we have that $$ \lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}=1 $$