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Does this the sequence go to zero?

$\Pi_{n=1}^{N}\text{sin}(2\pi n\omega)$ as N $\rightarrow \infty$ for any $\omega \in (0,1)?$

I can see this sequence is always decreasing for general $\omega$. And for some specific $\omega$ (in $\mathbb{Q}$ I believe), sin($2\pi n_0\omega$) becomes 0 for some specific $n_0$ so making the sequence to be 0 there after. But how to show such sequence goes to zero for general $\omega \in (0,1)$ though?

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  • $\begingroup$ See "help" above to find what sort of questions to ask here. $\endgroup$
    – Gerald Edgar
    Commented Oct 29, 2015 at 2:43
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    $\begingroup$ Hint: Kronecker's theorem for $\omega\not\in\mathbb Q$. $\endgroup$
    – Fan Zheng
    Commented Oct 29, 2015 at 3:12
  • $\begingroup$ Fan Zheng - what theorem do you mean exactly? $\endgroup$
    – Sergei
    Commented Oct 29, 2015 at 12:32

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