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. 



Is there a simple proof of the following Identity for $\sum_{k=m-1}^l(-1)^{k+m}\frac{k+2}{k+1}{\binom l k}\binom{k+1}m$?

> 
> We obtain for $l,m\in\mathbb{N}$ with $0\leq m-1 \leq l$:
>  \begin{align\*}
> \color{blue}{\sum\_{k=m-1}^{l}}&\color{blue}{(-1)^{k+m}\frac{k+2}{k+1}\binom{l}{k}\binom{k+1}{m}}\\
> &=\frac{1}{m}\sum\_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l}{k}\binom{k}{m-1}\tag{1}\\
> &=\frac{1}{m}\binom{l}{m-1}\sum\_{k=m-1}^l(-1)^{k+m}(k+2)\binom{l-m+1}{k-m+1}\tag{2}\\
> &=\frac{1}{m}\binom{l}{m-1}\sum\_{k=0}^{l-m+1}(-1)^{k+1}(k+m+1)\binom{l-m+1}{k}\tag{3}\\
> &=\frac{m+1}{m}\binom{l}{m-1}\sum\_{k=0}^{l-m+1}(-1)^{k+1}\binom{l-m+1}{k}\\
> &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum\_{k=1}^{l-m+1}(-1)^{k+1}\binom{l-m}{k-1}\tag{4}\\
> &=-\frac{m}{m+1}\binom{l}{m-1}[[l=m-1]]\\
> &\qquad+\frac{l-m+1}{m}\binom{l}{m-1}\sum\_{k=0}^{l-m}(-1)^k\binom{l-m}{k}\tag{5}\\
> &=\left(-1-\frac{1}{m}\right)[[l=m-1]]+\frac{l-m+1}{m}\binom{l}{m-1}[[l=m]]\tag{6}\\
> &\color{blue}{=\left(-1-\frac{1}{m}\right)[[l=m-1]]+1[[l=m]]}
> \end{align\*}
>  and the claim follows.
> 
> 
> 


*Comment:* 


* In (1) we use the binomial identity
$$\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$$
* In (2) we use the binomial identity
$$\binom{p}{q}\binom{q}{r}=\binom{p}{r}\binom{p-r}{q-r}$$
* In (3) we shift the index of the sum to start with $k=0$.
* In (4) we split the sum and work similarly as in (1).
* In (5) we do some simplifications regarding $(1-1)^{l-m+1}$ using *Iverson brackets*. We also shift the index of the sum again.
* In (6) we do a similar job as in (5).


