While studying the behaviour of umbilic points on Weingarten surfaces I discovered that the following combinatorial identity must be true.

For all $l,m\in{\mathbb N}$ with $l\geq m-1\geq0$ the following holds: $ \sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{l \choose k}{k+1 \choose m}= \left\{\begin{array}{ccl} 0&if& l>m\\ 1&if& l=m\\ -1-\frac{1}{l+1} &if& l=m-1 \end{array} \right. $

Improbable as it at first appears, it is easy to check the second and third options are true, and I have computer-checked and found it is true for all $l,m\leq 60$. Perhaps it is well-known or has an easy proof. Either would be good to know.