<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] 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 $(n+k)! = (n+k)(n+k-1)\cdots (n+1)n!$
</s>