The convergence of $(n\sin n)$ I  think $\lim_{n\to \infty} n\sin n$ does not exist, for $\sin x$ is a wave function and when $n\to \infty$, then so does factor $n$. But I am confused that how to give a strict proof for the divergence.
I would appreciate it if someone could give some suggestions and comments. 
 A: Every convergent sequence is bounded.
Edit: Let $m \in \mathbb{Z}.$ Then $$\sin x\geq \sin\left(\frac12\right)>0\,\, \forall x \in \left[\frac12+2m\pi,\frac32+2m\pi\right].$$ Moreover there exists $n_m \in \left[\frac12+2m\pi,\frac32+2m\pi\right] \cap \mathbb{N}.$ Therefore $$n_m\sin(n_m)\geq \left(\frac12+2m\pi\right)\sin\left(\frac12\right).$$It follows that $(n \sin n)_n$ is unbounded.
A: Given that $$\lim_{n \to \infty} \frac{1 }{n} = 0$$ And $\lim_{n \to \infty} \sin n$ doesn't exist, it's easy to prove that. The former can be proved by the definition while the latter with the proof by contradiction. Suppose  $$ \lim _{n \to \infty}n\sin n = L$$ So we have $$ \lim _{n \to \infty}n\sin n \times \frac{1}{n}= \lim _{n \to \infty} \sin n = 0$$
And this is a contradiction.
Edit: If we want to prove that $\lim_{n \to \infty} \sin n$ doesn't exist, according to the Jonas Teuwen's answer in the mentioned page, suppose $\lim_{n \to \infty} \sin n = l$. It's known that a sequence converges iff every subsequence converges. So $\lim_{n \to \infty} \sin 2n = \lim_{n \to \infty} \sin 2(n+1) = l$ and $\lim_{n \to \infty} \cos2n = \lim_{n \to \infty} 1-2\sin^2 n = 1 - 2l^2$ and also $\lim_{n \to \infty} \cos2(n+1) = 1 - 2l^2$. Then $$\sin(2) = \sin(2(n+1) - 2n) = \sin(2(n+1))\cos(2n) - \sin(2n)\cos(2(n+1))$$Taking limit yields $$\lim_{n \to \infty} \sin(2) = l\times (1-2l^2) - l\times (1-2l^2) = 0$$So this implies $$\sin(2) = 0$$ Which is clearly a contradiction.
A: We have that $a_n = n \sin{(n)}$. Rewrite it during that $\sin{(x)}\in\left[-1,1\right]$:
$$-n\leq n \sin{(n)} \leq n$$
Than, using the lim:
$$-lim_{n\rightarrow\infty}n \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq lim_{n\rightarrow\infty}n$$
$$-\infty \leq lim_{n\rightarrow\infty}n \sin{(n)} \leq \infty$$
This shows, that $a_n$ is unbounded and thats because limit does not exists:
$$\nexists lim_{n\rightarrow\infty} n \sin{(n)}$$
