A specific binomial summation identity Let $m$ and $n$ be positive integers with $m < n$. Prove
\begin{equation}
   \left(\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{n-k} \right)\left(\sum_{k=0}^m \binom{n}{k}\frac{(-1)^k}{k+1} \right) =\sum_{k=0}^m \binom{m}{k}\frac{(-1)^k}{(n-k)(k+1)} 
\end{equation}
I'm not sure how to approach this problem. A Hint would be greatly appreciated!
 A: We seek to prove for $n,m\ge 1$ with $m\lt n$ that
$$\sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{(n-k)(k+1)}.$$
The RHS is
$$\frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}
+ \frac{1}{n+1} \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{k+1}.$$
The second term is
$$\frac{1}{n+1} \frac{1}{m+1} \sum_{k=0}^m
{m+1\choose k+1} (-1)^k
= \frac{1}{n+1} \frac{1}{m+1}.$$
This means we evidently require
$$S_{m, n} = \sum_{k=0}^m {m\choose k} \frac{(-1)^k}{n-k}.$$
With this in mind we introduce
$$f(z) = m! (-1)^m \frac{1}{n-z} \prod_{q=0}^m \frac{1}{z-q}.$$
This has the property that for $0\le k\le m$
$$\mathrm{Res}_{z=k} f(z)
= m! (-1)^m \frac{1}{n-k} \prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^m \frac{1}{k-q}
\\ = m! (-1)^m \frac{1}{n-k}
\frac{1}{k!} \frac{(-1)^{m-k}}{(m-k)!}
= {m\choose k} \frac{(-1)^k}{n-k}.$$
Now with residues summing to zero and the residue at infinity being
zero we get
$$S_{m,n} = -\mathrm{Res}_{z=n} f(z)
= m! (-1)^m \prod_{q=0}^m \frac{1}{n-q}
\\ = (-1)^m \frac{m! \times (n-m-1)!}{n!}
= \frac{(-1)^m}{m+1} {n\choose m+1}^{-1}.$$
This means we need to show
$$\frac{(-1)^m}{m+1} {n\choose m+1}^{-1}
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \frac{1}{n+1} \frac{(-1)^m}{m+1} {n\choose m+1}^{-1}
+ \frac{1}{n+1} \frac{1}{m+1}$$
which is
$$(-1)^m
\sum_{k=0}^m {n\choose k} \frac{(-1)^k}{k+1}
= \frac{1}{n+1} (-1)^m
+ \frac{1}{n+1} {n\choose m+1}.$$
Note however that the LHS is
$$\frac{(-1)^m}{n+1} \sum_{k=0}^m {n+1\choose k+1} (-1)^k
= \frac{1}{n+1} (-1)^m +
\frac{1}{n+1} (-1)^m \sum_{k=-1}^m {n+1\choose k+1} (-1)^k
\\ = \frac{1}{n+1} (-1)^m
- \frac{1}{n+1} (-1)^m \sum_{k=0}^{m+1} {n+1\choose k} (-1)^k.$$
Working with the sum we find
$$[z^{m+1}] \frac{1}{1-z}
\sum_{k\ge 0} {n+1\choose k} z^k (-1)^k
= [z^{m+1}] \frac{1}{1-z}
(1-z)^{n+1}
\\ = (-1)^{m+1} {n\choose m+1}.$$
We finally obtain
$$\frac{1}{n+1} (-1)^m 
- \frac{1}{n+1} (-1)^m
(-1)^{m+1} {n\choose m+1}
= \frac{1}{n+1} (-1)^m
+  \frac{1}{n+1} {n\choose m+1}$$
which is the RHS as desired.
