A polynomial $P$ of degree $n \in \mathbb{N}$ satisfies:

$$
\text { for } k=0,1, \ldots, n, P(k)=\frac{k}{k+1}
$$

Determine $P(n+1)$.

Hint: Consider looking at $(X+1) P$.
Let $Q$ be the polynomial defined by: $Q=(X+1) P(X)-X$. $Q$ is of degree $n+1$ and has $0,1, \ldots n$ as roots, so $Q=a X(X-1) \ldots(X-n)$ where $a \in \mathbb{R}$. Now, $(n+2) P(n+1)-n-1=$ $Q(n+1)=a \times(n+1)$! Therefore, $P(n+1)=\frac{(n+1)(1+a \times n!)}{n+2}$. To determine $a$, for example: $0+1=Q(-1)=(-1)^{n+1} a(n+1)$!, i.e., $a=\frac{(-1)^{n+1}}{(n+1)!}$ and $P(n+1)=\frac{n+1+(-1)^{n+1}}{n+2}=$ 1 or $\frac{n}{n+2}$.

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