TITLE: Limit $\lim_{n \to \infty }\frac{n!}{(n+1)!}\frac{\sin((n+1)x)}{\sin(nx)}$
QUESTION [5 upvotes]: I am trying to evaluate this limit:
$$ \lim_{n \to \infty }\frac{n!}{(n+1)!}\frac{\sin((n+1)x)}{\sin(nx)}$$
My thought was to split it up as follows:
$$ = \lim_{n \to \infty }\frac{n!}{(n+1)!}\cdot\lim_{n \to \infty }\frac{\sin((n+1)x)}{\sin(nx)}
$$
$$
= 0\cdot \lim_{n \to \infty}\frac{\sin((n+1)x)}{\sin(nx)}=?$$
I am having difficulty continuing from here. Can anyone explain how this limit equals zero?

REPLY [4 votes]: As has been noted in the comments, $\lim_{n\to\infty}\frac{\sin((n+1)x)}{\sin(n x)}$ depends on $x$ and in general doesn't exist. Thus your approach to split up the limits is invalid.
However, I'm going to go out on a limb and say that you may have been trying to use the Ratio Test to determine the convergence of
$
\sum_{n=0}^{\infty}\frac{\sin(n x)}{n!}
$. If this is the case, you can get away with a much stronger result while doing much less work. The series is absolutely convergent since $|\sin(\theta)|\le 1$ for real $\theta$, and $\sum_{n=0}^{\infty} \frac{1}{n!}$ converges. In fact, we have
$$
\sum_{n=0}^{\infty}\frac{\sin(n x)}{n!} = \Im\left(\sum_{n=0}^{\infty}\frac{(e^{ix})^n}{n!}\right)
$$
$$
=\Im\left(e^{e^{ix}}\right)= e^{\cos(x)}\sin(\sin(x))
$$