Finding $\lim\limits_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$ Problem: Find $\lim_{n \to \infty}(\sin(1)\cdot \sin(2) \cdot \ldots \cdot \sin(n))$.
My idea: I suppose it to be $0$, but how might I go about proving this?

First off observe that $|\sin n| \leq 1$ for all $n$. Then can you show that there are infinitely many $n$ which are a bounded distance from 1 (i.e. at most equal to $1 - \epsilon$ for some $\epsilon > 0$)?
Then it follows that $$\prod_{n \in \mathbb N} |\sin n \,| \leq \prod_{n \in \mathbb{N}} (1 - \epsilon) = \lim_{n \rightarrow \infty} (1-\epsilon)^n$$
