Let $n \geqslant 1$ and $P \in \mathbb{R}_{n}[X]$ such that $\forall k \in \llbracket 0, n \rrbracket, P(k)=\frac{k}{k+1}$ Calculate $P(n+1)$

Theorem 1 (Viète's Formulas). Let $P=X^{2}-s X+p$ and $\alpha, \beta$ be its roots. Then, $s=\alpha+\beta$ and $p=\alpha \beta$.

Let $P=X^{3}-s X^{2}+t X-p$ and $\alpha, \beta, \gamma$ be its roots. Then $s=\alpha+\beta+\gamma, t=\alpha \beta+\alpha \gamma+\beta \gamma$ and $p=\alpha \beta \gamma$

Proof. $P=(X-\alpha)(X-\beta)=X^{2}-(\alpha+\beta) X+\alpha \beta$. By identification, we retrieve the indicated formulas. We proceed similarly for degree 3.
The given relation is not immediately polynomial, so it needs to be transformed. \( P(k) = \frac{k}{k+1} \Longleftrightarrow (k+1) P(k) = k \Longleftrightarrow (k+1) P(k) - k = 0 \). Thus, the polynomial \((X+1) P(X) - X\) has \(n+1\) roots. Since the degree of \((X+1) P(X)\) is strictly greater than 1, \((X+1) P(X) - X\) is of degree \(n+1\). Similarly as before, \((X+1) P(X) - X = c X(X-1) \ldots (X-n)\). It remains to determine \(c\). To compute the left-hand side, we need to cancel \((X+1) P(X)\) because we know nothing useful about \(P\) and thus evaluate at \(-1: 1 = c \cdot (-1)^n \cdot (n+1)! \Longleftrightarrow c = \frac{(-1)^n}{(n+1)!}\)

Finally, \(P(n+1) = \frac{n+1 + (-1)^n}{n+2}\)