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

This is a recent exam question, which I couldn't figure out in the exam. My guess is it doesn't exist. 
 A: 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
$$\lim_{n\to\infty}\,n\,\sin\frac1n=\lim_{n\to\infty}\frac{\sin\frac1n}{\frac1n}=1\;\;\ldots\ldots$$
A: For small $|x|$ we have $\sin(x)\approx x$.
Hence 
\[  i \cdot \sin\frac{1}{i} \approx 1  \]
for big values of $i$. 
Hence 
\[ \frac{\sum_{i=1}^n i\cdot \sin\frac{1}{i} }{n}\approx \frac{n}{n}=1\]
A: 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.
A: 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
$$
\lim_{n\to\infty} \sum_{i=1}^n \frac{i\sin(1/i)}{n}=1
$$
