<s>[INST] $a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$, $\lim_{n\to \infty}a_n$ is Given $a_1,a_2,n\in \mathbb N$
$$a_n=(1-\frac{1}{n})a_{n-1}+\frac{1}{n}a_{n-2}$$
Then $\lim_{n\to \infty}a_n$ is
(A) $2(a_2-a_1)+a_1e^{-1}$
(B) $2(a_1-a_2)e^{-1}+a_2$
(C) $2(a_1-a_2)e^{-1}+a_1$
(D) $2(a_2-a_1)e^{-1}+a_1$
My attempt,
$a_1,a_2\in \mathbb N$
$$a_3=(1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1}$$
$$a_4=(1-\frac{1}{4})a_{3}+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})+\frac{1}{4}a_{2}$$ $$=(1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1}$$
$$a_5=(1-\frac{1}{5})a_{4}+\frac{1}{5}a_{3}$$ $$=(1-\frac{1}{5})((1-\frac{1}{4})(1-\frac{1}{3})a_{2}+(1-\frac{1}{4})\frac{1}{3}a_{1})+\frac{1}{5}((1-\frac{1}{3})a_{2}+\frac{1}{3}a_{1})$$ 
But I am not able to conclude from here, I couldn't able to generalise anything from here. 
 [/INST] Multiplying both sides of the equation by $n$ and rearranging, you get that: $$n(a_n-a_{n-1}) = -(a_{n-1}-a_{n-2})$$
Setting $b_n = a_n-a_{n-1}, n\geq 2$ and $b_1 =a_1$, this becomes: $$b_n = \frac{-1}{n}b_{n-1} = ... = (-1)^n\frac{2}{n!}b_2 =  (-1)^n\frac{2}{n!}(a_2-a_1),\text{ for } n \geq2.$$
So, $a_n = a_n-a_{n-1} + a_{n-1} - a_{n-2}+...-a_1+a_1 = \sum \limits_{i=1}^n b_i = a_1 + 2(a_2-a_1)\sum \limits_{i=2}^n \frac{(-1)^n}{n!}.$
It remains to calculate $\lim\limits_{n\rightarrow \infty}a_n = a_1+2(a_2-a_1)\sum \limits_{i=2}^\infty \frac{(-1)^n}{n!}.$ But this infinite series is the Taylor expansion of $e^x$ evaluated at $x=-1$.
So, the final answer is: $$\lim\limits_{n\rightarrow \infty}a_n =  a_1+2(a_2-a_1)e^{-1}.$$
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