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How can I prove that this sequence does not converge, using the definition?

$$W_n = \sin(n^3)$$

For $n \in \mathbb{N}$. I tried to do a proof by reduction to the absurd but without result.

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    There is no limit, the sequence do not converge to anywhere. – 2017-02-25
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    I know that it does not converge but i don't know how to prove it. – 2017-02-25
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    We say that a sequence _diverges_ if it goes to plus or minus infinity. If a sequence do not converge we say that the _limit do not exist_ I think someone will answer this but as a felling you could try to (1) show that two different sub-sequences converge to different values. – 2017-02-25
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    I couldn't find any... The n^3 bother me. – 2017-02-26
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    (+1) In hope that someone answer it. Seems a very technical prove – 2017-02-26
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    @RafaelWagner I'm pretty sure that's not true. Any sequence that does not converge is said to diverge, even if it doesn't go to plus or minus infinity. – 2017-02-26

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Reference: https://terrytao.wordpress.com/2010/03/28/254b-notes-1-equidistribution-of-polynomial-sequences-in-torii/

As a special case of Corollary 6, we have

Let $P(n)=a_s n^s + \cdots + a_0$ be a polynomial with real coefficients. If $a_s$ is irrational, then $P(n)$ mod $1$ is equidistributed.

With $P(n)=\frac1{2\pi} n^3$, we have $P(n)$ mod $1$ is equidistributed. Thus, $n^3$ mod $2\pi$ is equidistributed. Then it follows that $\sin n^3$ is dense in $[-1,1]$.