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
 A: 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}
A: $$n(n+1)a_{n+1}=n(n-1)a_n-(n-2)a_{n-1}$$
Multiplying both sides by $(n-1)!$, we get,
$$(n+1)!a_{n+1}=n!(n-1)a_n-(n-1)!(n-2)a_{n-1}$$
Now, denoting $b_n:=n!\cdot a_n$, we rewrite the above as,
$$b_{n+1}=(n-1)b_n-(n-2)b_{n-1}$$
Rewriting,
$$b_{n+1}-b_n=n(b_n-b_{n-1})-2(b_n-b_{n-1})=(n-2)(b_n-b_{n-1})$$
Denoting $c_n:=b_{n+1}-b_n$, we have,
$$c_n=(n-2)c_{n-1}\implies c_n=(n-2)!c_2$$
But $c_2=(2-2)c_1=0$ and hence $c_n=0$ for all $n$ which implies $b_{n+1}=b_n$ for all $n$, i.e., the sequence $(b_n)$ is a constant, so $b_n=b_0=a_0$ and thus $a_n=\dfrac{b_n}{n!}=\dfrac{a_0}{n!}$

It's now obvious that the required limit is $0$ since $a_0$ is a constant while $n!$ grows unbounded as $n\to\infty$, so $\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}\dfrac{a_0}{n!}= 0$
