Let $n \in \mathbb{N}^{*}$ and $P \in \mathbb{R}[X]$ of degree at most $n$ such that $P(k)=\frac{k}{k+1}$ for all $k \in \llbracket 0, n \rrbracket$. What is the value of $P(n+1)$?
Let $Q(X)=(X+1) P(X)-X$ : the polynomial $Q$ is of degree at most $n+1$ and we have $Q(k)=0$ for all $k \in\{0,1, \ldots, n\}$. It is therefore of the form:

$$
Q(X)=c X(X-1)(X-2) \ldots(X-n)
$$

Moreover, we have $Q(-1)=1$ so $c=\frac{1}{(-1)(-2) \ldots(-n-1)}=\frac{(-1)^{n+1}}{(n+1)!}$, thus

$$
P(X)=\frac{1}{X+1}\left((-1)^{n+1} \frac{X(X-1)(X-2) \ldots(X-n)}{(n+1)!}+X\right)
$$

We obtain $P(n+1)=\frac{n+1+(-1)^{n+1}}{n+2}$.