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.
Every convergent sequence is bounded.

Edit: Let $m \in \mathbb{Z}.$ Then $$\sin x\geq \sin\left(\frac12\right)&gt;0\,\, \forall x \in \left[\frac12+2m\pi,\frac32+2m\pi\right].$$ 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.
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:
$$\nexists lim_{n\rightarrow\infty} n \sin{(n)}$$