If a degree $n$ polynomial has $f(k)=k/(k+1)$ for $k=0,1,\ldots,n$, what is $f(n+1)$?

Question

Suppose $f(x)\in K[x]$, $\deg f(x)=n$ and $f(k)=k/(k+1)$ for any $k=0,1,2,\ldots ,n$, what is $f(n+1)$?

I have a vague memory that there is a very clever trick that can solve this problem easily, but I simply can't recall how.

I have tried using brutal force, applying Lagrange interpolation formula. But the result looks very ugly (a sum that I don't know how to reduce). Certainly that's not the desirable approach here.

Answer 1

Hint

Let $g(x)=(x+1)f(x)-x$. This is a polynomial with degree $n+1$ and you know its roots, namely $x=0,1,2, \ldots ,n$. Thus $$g(x)=Ax(x-1)(x-2) \dotsb (x-n),$$ where $A$ can be determined to be $A=\frac{(-1)^{n+1}}{(n+1)!}$