Ratio test involving factorials: $a_{n} = \frac{n-2}{(n+1)!}$; finding $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$

Question

The correct answer is apparently $0$ however I end up with $1$. I do the following:

$$a_{n+1} = \frac{(n+1)-2}{((n+1)+1)!} = \frac{n-1}{(n+2)!}$$

$$ \begin{split} \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n \to \infty} \left|\frac{n-1}{(n+2)(n!)} \times \frac{(n+1)(n!)}{n-2}\right| \\ &= \lim_{n \to \infty} \left|\frac{(n-1)(n+1)}{(n+2)(n-2)}\right| \\ &= \lim_{n \to \infty} \left|\frac{n^2-1}{n^2-2}\right| \\ &= \lim_{n \to \infty} \frac{n^2}{n^2} \\ &= 1 \end{split} $$

I assumed as $n$ approaches $\infty$ the '$-1$' and '$-2$' become meaningless, however I'm pretty certain that is now incorrect unless I've made an error earlier on.

Answer 1

Note that $(n+2)! = ((n+1)!)(n+2)$

I think that is the mistake.

$$\lim_{n \rightarrow \infty} \frac{n-1}{(n+1)!(n+2)} \frac{(n+1)!}{n-2}=\lim_{n \rightarrow \infty} \frac{n-1}{(n+2)(n-2)}=0$$

Answer 2

It should be $$\begin{align*}\lim_{n \to \infty} \left|\frac{n-1}{(n+2)(n+1)n!} \cdot \frac{(n+1)n!}{n-2}\right| = \lim_{n\to \infty} \frac{n-1}{(n+2)(n-2)} = 0\end{align*}$$

since $$\begin{align*}(n+2)! &= (n+2)(n+1)\underbrace{\color{blue}{(n)(n-1)\cdots (2)(1)}}_{\color{blue}{n!}} \\ & = (n+2)(n+1)\color{blue}{n!}\end{align*}$$

In general, we have $(n+k)! = (n+k)(n+k-1)\cdots (n+1)n!$