<s>[INST] Ratio test involving factorials: $a_{n} = \frac{n-2}{(n+1)!}$; finding $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$ 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. 
 [/INST] Note that $(n+2)! = ((n+1)!)(n+2)$
I think that is the mistake.
$$\lim_{n \rightarrow \infty} \frac{n-1}{(n+1)!(n+2)} \frac{(n+1)!}{n-2}=\lim_{n \rightarrow \infty} \frac{n-1}{(n+2)(n-2)}=0$$
</s>