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?

Consider that in every interval of the form $[\pi n-\frac{1}2,\pi n+\frac{1}2]$ there is at least one integer. Call the set of integers within such an interval $S$ and note that, as there are infinitely many such (disjoint) intervals, $S$ is infinite. However, in such intervals $|\sin(x)|<\sin(\frac{1}2)<1$. If we let $r=\sin(\frac{1}2)$, this, together with the fact that $|\sin(x)|\leq 1$, it follows that a product including the sines of $k$ members of $S$ has absolute value at most $r^k$. Since $k$ goes to infinity, the products can be constrained to any neighborhood of $0$, hence the product converges to $0$.
One can easily make this argument as rigorous as they desire.
