Limit $\lim_{n \to \infty }\frac{n!}{(n+1)!}\frac{\sin((n+1)x)}{\sin(nx)}$

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?

Solution 1:

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 $$ \sum_{n=0}^{\infty}\frac{\sin(n x)}{n!} = \Im\left(\sum_{n=0}^{\infty}\frac{(e^{ix})^n}{n!}\right) $$ $$ =\Im\left(e^{e^{ix}}\right)= e^{\cos(x)}\sin(\sin(x)) $$