Convergence of finite product of sine functions

Question

How can I prove that the finite product $$ a_{n} = \sin(1) \sin(\frac{1}{2}) \sin(\frac{1}{3})...\sin(\frac{1}{n})$$ converges or not?

Answer 1

If and only if $x$ is positive, we can say that $\sin(x)$ is less than $x$. We can then use the squeeze theorem to say:

$\lim_{x \rightarrow \infty} 0 < \lim_{x \rightarrow \infty} \sin(\frac{1}{x}) < \lim_{x \rightarrow \infty} \frac{1}{n!}$

Which yields zero.

Answer 2

For positive $x$, we have that $\sin(x) < x$. Then we have the following inequalities:

$$0 < \sin(1)\cdots\sin\left(\frac{1}{n}\right) < \frac{1}{n!}$$

So the product goes to $0$ by squeeze theorem.