19. Given that $f(x)$ is an $n$-degree polynomial, and $f(k)$
$$
=\frac{k}{k+1}, k=0,1, \cdots, n \text {, }
$$

find $f(n+1)$.
19. Analysis: $g(x)=(x+1) f(x)-x$ is a polynomial of degree $n+1$, and by (1), $g(x)$ is 0 when $x=0,1, \cdots, n$. Therefore, $g(x)=a x(x-1) \cdots(x-n)$, where $a$ is a constant to be determined.

Let $x=-1$, we get $1=g(-1)=a \cdot(-1)^{n+1} \cdot$ $(n+1)!$,

i.e., $a=\frac{(-1)^{n+1}}{(n+1)!}$, so $(n+2) f(n+1)-(n$ $+1)=g(n+1)=(-1)^{n+1}$,
$$
f(n+1)=\frac{n+1+(-1)^{n+1}}{n+2} .
$$