Combinatorics sum to 1 using Identity I want to prove that the following equals 1:
$W=\sum_{i=0}^{n-r}(-1)^i n \frac{C_{n-1}^{r-1}C_{n-r}^i}{(r+i)}$. I tried mathematical induction and succeeded. Is there any known identity of combinatorics that could be used to prove it directly?
 A: If  the question asked is (i.e., the first fraction inside the sum is $\frac{n}{r}$ instead of $\frac{n}{r+1}$) as follows, it makes sense. 
\begin{eqnarray*}
 \sum_{i+r=n}{ \frac{n}{r}  \binom{n-1}{r-1} \binom{n-r}{i}  (-1)^{i}}  &=& 1\\
\end{eqnarray*}
This comes straightforward from multinomial expansion.
\begin{eqnarray*}
\left(x_{1}+x_{2}+x_{3} \right)^{n} &=& \sum_{i+r=n}{\binom{n}{r,i} x_{1}^{i} x_{2}^{r}x_{3}^{n-r-i}} \\
\end{eqnarray*}
where 
$\binom{n}{r,i}  =\frac{n!}{r! i! (n-r-i)}$
Choose $x_1=-1,x_{2}=x_{3}=1$, we will have, 
\begin{eqnarray*}
\left(-1+1+1 \right)^{n} &=& \sum_{i+r=n}{\binom{n}{r,i} \left(-1\right)^{i} } \\
 &=& \sum_{i+r=n}{ \frac{n}{r}  \binom{n-1}{r-1} \binom{n-r}{i}  (-1)^{i}} \\
\end{eqnarray*}
Note that, 
\begin{eqnarray*}
\frac{n}{r}  \binom{n-1}{r-1} \binom{n-r}{i} &=& \frac{n}{r} \frac{(n-1)!}{(r-1)! (n-r)!}  \frac{(n-r)!}{(i!)(n-r-i)!} \\
&=&  \frac{(n)!}{(r)! (i!)(n-r-i)!} \\
&=& \binom{n}{r,i}
\end{eqnarray*}
A: We show the validity of OPs claim 
\begin{align*}
\sum_{i=0}^{n-r}(-1)^i\frac{n}{r+i}\binom{n-1}{r-1}\binom{n-r}{i}=1\qquad\qquad (0\leq r\leq n)
\end{align*}

We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{n-r}}&\color{blue}{(-1)^i\frac{n}{r+i}\binom{n-1}{r-1}\binom{n-r}{i}}\\
&=n\binom{n-1}{r-1}\sum_{i=0}^{n-r}\frac{(-1)^i}{r+i}\binom{n-r}{i}\tag{1}\\
&=n\binom{n-1}{r-1}\sum_{i=0}^{n-r}(-1)^i\binom{n-r}{i}\int_0^1x^{r+i-1}\,dx\tag{2}\\
&=n\binom{n-1}{r-1}\int_{0}^1x^{r-1}\sum_{i=0}^{n-r}(-1)^i\binom{n-r}{i}x^i\,dx\tag{3}\\
&=n\binom{n-1}{r-1}\int_0^1x^{r-1}(1-x)^{n-r}\,dx\tag{4}\\
&=\binom{n-1}{r-1}\binom{n-1}{r-1}^{-1}\tag{5}\\
&\,\,\color{blue}{=1}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we factor out all terms independent from $i$.

*In (2) we use $\int_0^1x^{k}\,dx=\frac{1}{k+1}$ with $k\ne -1$.

*In (3) we do some rearrangements as preparation for the next step.

*In (4) we apply the binomial theorem.

*In (5) we apply the beta function identity
$$
\binom{n}{r}^{-1}=(n+1)\int_0^1x^r(1-x)^{n-r}\,dx
$$
A: A Slightly More General Identity
$$
\begin{align}
\sum_k(-1)^k\frac{\binom{n}{k}}{\binom{r+k}{m}}
&=\sum_k(-1)^k\frac{\binom{n-1}{k}+\binom{n-1}{k-1}}{\binom{r+k}{m}}\tag1\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac1{\binom{r+k}{m}}-\frac1{\binom{r+k+1}{m}}\right)\tag2\\
&=\sum_k(-1)^k\binom{n-1}{k}\left(\frac{\frac{r+k+1}{m+1}}{\binom{r+k+1}{m+1}}-\frac{\frac{r+k+1-m}{m+1}}{\binom{r+k+1}{m+1}}\right)\tag3\\
&=\frac{m}{m+1}\sum_k(-1)^k\frac{\binom{n-1}{k}}{\binom{r+k+1}{m+1}}\tag4\\
&=\frac{m}{m+n}\frac1{\binom{r+n}{m+n}}\tag5
\end{align}
$$
Explanation:
$(1)$: Pascal's Rule
$(2)$: substitute $k\mapsto k+1$ in $(-1)^k\left.\binom{n-1}{k-1}\middle/\binom{r+k}{m}\right.$
$(3)$: put things over a common denominator
$(4)$: subtract and pull out the factor of $\frac{m}{m+1}$
$(5)$: repeat $n$ times

Application to the Question
$$
\begin{align}
\sum_{k=0}^{n-r}(-1)^kn\frac{\binom{n-1}{r-1}\binom{n-r}k}{r+k}
&=n\binom{n-1}{r-1}\sum_{k=0}^{n-r}(-1)^k\frac{\binom{n-r}k}{\binom{r+k}{1}}\\
&=n\binom{n-1}{r-1}\frac1{n-r+1}\frac1{\binom{n}{n-r+1}}\\
&=n\binom{n-1}{r-1}\frac1n\frac1{\binom{n-1}{n-r}}\\[6pt]
&=1
\end{align}
$$
