<s>[INST] If $n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$ >Then $\lim_{n\rightarrow \infty}a_{n}$ 
Let $a_{0}=1,a_{1}=2$ and for $n\geq 1\;, n(n+1)a_{n+1}=n(n-1)a_{n}-(n-2)a_{n-1}.$
Then $\lim_{n\rightarrow \infty}a_{n}$

$\bf{Attempt:}$ $2a_{2} = a_{0} = 1$ so $\displaystyle a_{2} = \frac{1}{2}$
$\displaystyle 6a_{3}=2a_{2} = 1$ so $\displaystyle a_{3} = \frac{1}{6}$
could some help me to solve it, thanks
 [/INST] Working through the first few $n$ values it can be seen that, for $n \geq 2$,
$$a_{n} = \frac{a_{0}}{n!}.$$
This yields the limit
$$\lim_{n \to \infty} a_{n} = \lim_{n \to \infty} \frac{a_{0}}{n!} = 0.$$
For verification:
\begin{align}
\frac{n(n+1) a_{n+1} - (n-1) a_{n}}{(n+1)(n+2)} &= \frac{a_{0} \, (n - (n-1))}{(n+2)!} = \frac{a_{0}}{(n+2)!} = a_{n+2}. 
\end{align}
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