How to solve $\lim_{n\to \infty}\sin(1)\times \sin(2)\times\sin(3)\times\ldots\times\sin(n)$

Solution 1:

For the first limit, it should be $0$ for the following reason: $\;\{\sin n\mid n\in \mathbf N\}$ is dense in $[-1,1]$. So for any $\varepsilon >0$ there exists $N$ such that $\lvert\, \sin N\,\rvert<\varepsilon$, which implies that for any $n\ge N$, $$\lvert\,\sin 1\cdot \sin 2\cdots \sin N\cdots\sin n\,\rvert <\varepsilon.$$

Solution 2:

Density (i.e., the irrationality of $\pi$) is not needed.

Let $f\colon\mathbb R$ be any periodic function with $|f(x)\le 1$ for all $x$ and there exists a closed interval $I$ of length $1$ and a number $q<1$ such that $|f(x)|\le q$ for all $x\in I$. Then for any $m\in\mathbb N_0$ $$\lim_{n\to \infty}n^mf(1)f(2)\cdots f(n) = 0 $$

To see this let $p$ be a period of $f$ Then $I, I+1, \ldots , I+\lceil p\rceil-1$ cover a full period of $f$, hence among any $\lceil p\rceil $ consecutive integers $k+i$, $0\le i<\lceil p\rceil$, there is at least one with $|f(k+i)\le q$. As $f(k)|\le 1$ for all other factors, we conclude $$\left|\prod_{k=1}^nf(k)\right| \le q^{\lfloor n/\lceil p\rceil\rfloor}<\frac1q\cdot(\sqrt[\lceil p\rceil]q)^n$$ This exponentially small bound implies the claim.

To apply this to the original problem with $f(x)=\sin x$ observe that one may take for example $I=[-\frac12,\frac12]$ and $q=\sin \frac12<\frac12$.