Combinatorial sum identity for a choose function $\sum\limits_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}$ I want to show that the following holds:
$$\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}.$$
I have an idea of what is going on here. On the RHS we are selecting $r+s+1$ elements from the set $[m+n+1]$. Another way of doing this is to partition the set $[m+n+1]$ into two sets in several ways by shifting $k$ elements from one set to the other. For example, we can partition $[m+n+1]$ into a set with $m+k$ elements and a set with $n+1-k$ elements. From the set with $m+k$ elements we can select $r$ elements and from the set with $n+1-k$ elements we can select $s+1$. However, this is not matching the LHS. I believe the problem is that we don't know which $k$ elements are being shifted around.
I thought rewriting the sum as follows would help:
$$\sum_{k=0}^{m+n} \binom{k}{r} \binom{n+m-k}{s}.$$
I just don't understand why the "$+1$" on the top and bottom of the RHS disappear. I have spent several hours on this problem, and I have run out of ideas. Please help, I truly appreciate it.

Suppose we seek to evaluate
$$\sum_{k=-m}^n {m+k\choose r} {n-k\choose s}
= \sum_{k=0}^{m+n} {k\choose r} {m+n-k\choose s}.$$
Here we may assume that $r$ and $s$ are non-negative. We introduce
$${m+n-k\choose s} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-k-s+1}} \frac{1}{(1-z)^{s+1}} \; dz.$$
This integral  controls the range, being  zero when $k\gt  m+n$ and we
may extend the range of $k$ to infinity. We get for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-s+1}} \frac{1}{(1-z)^{s+1}} 
\sum_{k\ge 0} {k\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-s+1}} \frac{1}{(1-z)^{s+1}} 
\sum_{k\ge r} {k\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-s+1}} \frac{z^r}{(1-z)^{s+1}} 
\sum_{k\ge 0} {k+r\choose r} z^k
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-s+1}} \frac{z^r}{(1-z)^{s+1}} 
\frac{1}{(1-z)^{r+1}}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1}{z^{m+n-r-s+1}} \frac{1}{(1-z)^{r+s+2}} 
\; dz.$$
We get for the answer
$${m+n-r-s+r+s+1\choose r+s+1}
= {m+n+1\choose r+s+1}.$$
