1
$\begingroup$

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?

$\endgroup$

closed as off-topic by Gerald Edgar, Noah Schweber, Myshkin, Alexey Ustinov, Todd Trimble Oct 29 '15 at 5:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "MathOverflow is for mathematicians to ask each other questions about their research. See Math.StackExchange to ask general questions in mathematics." – Gerald Edgar, Noah Schweber, Myshkin, Alexey Ustinov, Todd Trimble
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ See "help" above to find what sort of questions to ask here. $\endgroup$ – Gerald Edgar Oct 29 '15 at 2:43
  • 1
    $\begingroup$ Hint: Kronecker's theorem for $\omega\not\in\mathbb Q$. $\endgroup$ – Fan Zheng Oct 29 '15 at 3:12
  • $\begingroup$ Fan Zheng - what theorem do you mean exactly? $\endgroup$ – Sergei Oct 29 '15 at 12:32