A polynomial $p$ of degree $n$ takes the value $\frac{k}{k+1}$ at $k$ when $k=0,1, \ldots, n$. What value can $p(n+1)$ take?
Consider the polynomials $q(x)=(x+1) p(x)$ and

$$
r(x)=x+\frac{1}{(n+1)!}(0-x)(1-x) \ldots(n-x)
$$

It is clear that both are polynomials of degree at most $(n+1)$, and for $k=0,1, \ldots, n$, we have $q(k)=r(k)=k$. Additionally,

$$
r(-1)=-1+\frac{1}{(n+1)!}(0-(-1))(1-(-1)) \ldots(n-(-1))=0=q(-1)
$$

Thus, the values of the polynomials $q$ and $r$ coincide at $n+2$ distinct points. This implies that the two polynomials are identical, hence

$$
p(n+1)=\frac{q(n+1)}{n+2}=\frac{r(n+1)}{n+2}=\frac{n+1+(-1)^{n+1}}{n+2}
$$

Therefore, the value of $p(n+1)$ is 1 if $n$ is odd, and $\frac{n}{n+2}$ if $n$ is even.