Simplifying Sum How would one show that
$$
\sum_{i=0}^n\binom{n}{i}(-1)^i\frac{1}{m+i+1}=\frac{n!m!}{(n+m+1)!} ?
$$
Any hint would be appreciated.
Note: I tried to recognize some known formula, but since I don't have much knowledge in this area I failed to. I tried expanding the binomial coefficient and factoring $n!$, but still I can't get the result. I tried writing out each term to regroup them in a particular way or to see some pattern, e.g. telescoping, but it didn't lead me anywhere (and since it gets quite messy for nothing I didn't reproduce this here).
 A: You can prove it by induction on $n$. First observe that the right-hand side is 
$$\frac1{n+m+1}\binom{n+m}n^{-1}\;,$$
and multiply through by $\binom{n+m}n$. Now suppose as induction hypothesis that 
$$\sum_{i=0}^n\binom{n}i\binom{n+m}n(-1)^i\frac1{m+1+i}=\frac1{m+1+n}$$
for all $m\ge 0$. Then
$$\begin{align*}
\sum_{i=0}^{n+1}&\binom{n+1}i\binom{n+1+m}{n+1}(-1)^i\frac1{m+1+i}\\
&=\sum_{i=0}^{n+1}\left(\binom{n}i+\binom{n}{i-1}\right)\binom{n+1+m}m(-1)^i\frac1{m+1+i}\\
&=\frac{m+1+n}{n+1}\sum_{i=0}^n\binom{n}i\binom{n+m}m(-1)^i\frac1{m+1+i}\\
&\qquad+\sum_{i=0}^n\binom{n}i\binom{n+1+m}{m+1}\cdot\frac{m+1}{n+1}\cdot(-1)^{i+1}\frac1{m+2+i}\\
&\overset{(*)}=\frac1{n+1}-\frac{m+1}{n+1}\cdot\frac1{m+2+n}\\
&=\frac1{n+1}\left(1-\frac{m+1}{m+2+n}\right)\\
&=\frac1{m+2+n}\;,
\end{align*}$$
as desired, where the step $(*)$ applies the induction hypothesis both at $m$ and at $m+1$.
A: Suppose we seek to evaluate
$$S_m(n) = \sum_{p=0}^n {n\choose p} (-1)^p \frac{1}{m+p+1}$$
where $m$ is an integer not equal to $-1,-2,\ldots, -n-1$ where the
fraction is undefined.
Observe that with $$f(z) = \frac{(-1)^n n!}{z+m+1}
\prod_{q=0}^n \frac{1}{z-q}$$
this is
$$S_m(n) = \sum_{p=0}^n \mathrm{Res}_{z=p} f(z).$$
the reason being that
$$\mathrm{Res}_{z=p} f(z)
= \frac{(-1)^n n!}{m+p+1}
\prod_{q=0}^{p-1} \frac{1}{p-q} \prod_{q=p+1}^n \frac{1}{p-q}
\\ = \frac{(-1)^n n!}{m+p+1} \frac{1}{p!} \frac{(-1)^{n-p}}{(n-p)!}
= (-1)^p \frac{1}{m+p+1} {n\choose p}.$$
With $f(z)$ being rational the residues at all poles sum to zero and we thus have
$$S_m(n) = 
-(\mathrm{Res}_{z=-m-1} f(z) +\mathrm{Res}_{z=\infty} f(z)).$$
We get for the residue at infinity
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
\frac{(-1)^n n!}{1/z+m+1}
\prod_{q=0}^n \frac{1}{1/z-q}
= -\mathrm{Res}_{z=0} \frac{z^{n+2}}{z^2}
\frac{(-1)^n n!}{1+z(m+1)}
\prod_{q=0}^n \frac{1}{1-qz}
\\ = -\mathrm{Res}_{z=0} z^n
\frac{(-1)^n n!}{1+z(m+1)}
\prod_{q=0}^n \frac{1}{1-qz} = 0.$$
On the other hand the negative of the residue at $z=-m-1$ is
$$- (-1)^n n! \prod_{q=0}^n \frac{1}{-m-1-q}
= n! \prod_{q=0}^n \frac{1}{q+m+1}
= \frac{n! m!}{(n+m+1)!}.$$
This concludes the argument.
