Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k}{k+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
$Q(x) = (x+1)P(x)-x$ has degree $n+1$ and $n+1$ roots at $x=0$, $x=1$, through $x=n$, so it is of the form $Q(x) = ax(x-1)(x-2)\cdots (x-n)$. Moreover, $Q(-1) = 1$, and thus $\displaystyle a = (-1)^{n+1} \frac{1}{(n+1)!}$.
It follows that $Q(n+1) = (-1)^{n+1}$ and thus that $\displaystyle P(n+1) = \frac{n+1+(-1)^{n+1}}{n+2}$.
Hint: Consider the polynomial $Q(x)=(x+1)P(x)$. Then $Q(k)=k$.