The limit of $\sin(n^\alpha)$ (1) It is easy to prove that $\lim\limits_{n\to\infty}{\sin(n)}$ does not exist.
(2) I want to ask how to prove that $\lim\limits_{n\to\infty}{\sin(n^2)}$ does not exist.
(3) Furthermore, $\lim\limits_{n\to\infty}{\sin(n^k)}$ does not exist. ($k$ is a positive integer.)
(4) In addition, $\lim\limits_{n\to\infty}{\sin(n^{\alpha})}$ does not exist. ($\alpha$ is a positive real number.)
 A: Concepts I use here is "Weyl's Equidistribution Criterion"
Reference: http://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/
(1) The limit does not exist simply because the sequence $\{e^{in}\}$ is equidistributed on the unit circle. 
(2), (3) follows from Van der Corput's lemma:
 We can bring down exponent $k$-case to exponent $1$-case. 
 Better way to put it is: 
For any integer $k\geq 1$, the sequence $\{e^{in^{k}}\}$ is equidistributed on the unit circle. 
(4) For $0<\alpha<1$, the sequence $\{e^{in^{\alpha}}\}$ is dense in the unit circle, so the limit of $\sin(n^{\alpha})$ does not exist. 
For $\alpha>1$, and $\alpha=\frac{q}{p}\in\mathbb{Q}^{+}$, consider $\{\sin(m^p)^{\frac{q}{p}} \}$. This is a subsequence of $\{\sin (n^{\frac{q}{p}})\}$. 
The subsequence $\{\sin(m^q)\}$ diverges by (3). Thus the original sequence diverges as well. 
Remaining case is now $\alpha>1$ and irrational. 
This case can be settled by the lemma: 
(Lemma)
For a sequence of real numbers $\{x_n\}$, suppose that the set of all subsequential limits of $\{\exp(ix_n)\}$ is finite. Denote $D$ the difference sequence operator on the space of real sequences. Then for $\{y_n:=Dx_n=x_{n+1}-x_n\}$, we have also that the set of all subsequential limits of $\{\exp(iy_n)\}$ is finite. 
If $x_n=n^{\alpha}$, and $\lfloor\alpha\rfloor=m$, then $z_n:=D^m x_n \sim cn^{\alpha-m}$ for some positive constant $c$. Thus, by the same reason for the case $0<\alpha<1$, we have $\{\exp(iz_n)\}$ is dense in the unit circle. 
If $\{\sin(n^{\alpha})\}$ has a limit, then $\{\exp(ix_n)\}$ has a finite set of sequential limits. Then by lemma applied $m$-times, $\{\exp(iz_n)\}$ also has a finite set of sequential limits. This contradicts above. 
Hence, $\{\sin(n^{\alpha})\}$ does not have a limit.
A: We can prove that  $\exists \{c_i\}_{i=1}^{\infty} \subseteq \mathbb N$ and $c_1<c_2<c_3<...$ such that $\lim_{i\rightarrow \infty}\sin (c_i^{\alpha})=c$, $\forall c\in [-1,1]$,where $\alpha $ is a positive rational number. 
