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