The limit of $\sin(n^\alpha)$

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.