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

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.