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I was recently asked to find $$\liminf_{n->\infty} n|\sin(n)|. n \in \mathbb{N}$$

I suspect that the answer is $0$, but I'm not quite sure how to prove it. I know that $\{ \sin(n) : n \in \mathbb{N}\}$ is dense on the real line, but from this fact is not easy to me to deduce that $\{ n\sin(n) : n \in \mathbb{N}\}$ is also dense. Maybe one can use this inequality $$n |\sin n| \le |n^2-nk\pi|$$

for every $n,k \in \mathbb{N}$ and prove that $|n^2-nk\pi|$ can be as small as desired, but that seems hard to me.

mrprottolo
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    This is the same question as this: http://math.stackexchange.com/questions/1725321/liminf-limits-n-to-infty-n-sin-n. – PhoemueX Apr 03 '16 at 15:37
  • After looking at the related question, I believe that I must be wrong since I don't see anything deep going on. But what's wrong about finding a sequence like $(n_k|\sin(n_k)|)_{k\in\mathbb{N}}$ (where $n_k:=k\pi$ for example) that converges to $0$? Then $0$ must be a cluster point and since it's the smallest possible, we found our $\lim\inf$? Or are we exclusively looking at $n\in\mathbb{N}$ here? – Piwi Apr 03 '16 at 15:50
  • @Piwi I forgot to add that crucial hypothesis, edited. – mrprottolo Apr 03 '16 at 15:52

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