Consider the sequence $(k_n)$ defined by $k_{n+1}=n(k_n+k_{n-1})$ and $k_0=0,k_1=1$. What is $\lim_{n\to\infty}\frac{k_n}{n!}$?
My attempt: Consider the sequence $s_n=\frac{k_n}{n!}$ instead. Then, we have $s_{n+1}=\frac1{n+1}(ns_n+s_{n-1})=\frac n{n+1}s_n+\frac{s_{n-1}}{n+1}$, and our goal is to find the limit $\lim_{n\to\infty}s_n$.
How do we proceed from here?
The answer is given as $1-\frac1e$.
Taking the hint from the comments: $$s_{n+1}=\frac n{n+1}s_n+\frac1{n+1}s_{n-1}$$ $$s_{n+1}-s_n=-\frac1{n+1}(s_n-s_{n-1}) $$ $$=\frac{(-1)^n}{(n+1)!}(s_1-s_0)$$ $$=\frac{(-1)^n}{(n+1)!}$$ Then we have $$\lim_{n\to\infty}s_n=\lim_{n\to\infty}s_{n+1}-s_0=\sum_{k=0}^\infty s_{k+1}-s_k=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}=-\sum_{k=1}^\infty\frac{(-1)^k}{k!}$$ $$=-\left(-1+\sum_{k=0}^\infty\frac{(-1)^k}{k!}\right)=-\left(-1+\frac1e\right)=1-\frac1e$$