8. Given the sequence $\left\{x_{n}\right\}$, satisfying $(n+1) x_{n+1}=x_{n}$ $+n$, and $x_{1}=2$. Then $x_{200 s}=$ $\qquad$ .
8. $\frac{2005!+1}{2005!}$.

From $(n+1) x_{n+1}=x_{n}+n$, we derive $x_{n+1}-1=\frac{x_{n}-1}{n+1}$
Therefore, $x_{n+1}-1=\frac{x_{n}-1}{n+1}=\frac{x_{n-1}-1}{(n+1) n}$
$$
\begin{array}{l}
=\frac{x_{n-2}-1}{(n+1) n(n-1)}=\cdots \\
=\frac{x_{1}-1}{(n+1) n(n-1) \cdots 2}=\frac{1}{(n+1)!} .
\end{array}
$$

Hence $x_{n+1}=\frac{1}{(n+1)!}+1$.