An identity involving binomial coefficients Prove the following identity
$$\displaystyle \sum_{i+j=m}\frac{(n-1) \binom{ai+n-1}{i} \binom{aj+1}{j}}{(ai+n-1)(aj+1)} =  \frac{n\binom{am+n}{m}}{am+n}$$
where $i = 0,1,\cdots,m$ and $m, n$ are positive integers and $a$ is a positive integer or even a fraction.

One complete proof can be found in that famous book "Concrete Mathematics" by
Ronald L. Graham, Donald E. Knuth, and Oren Patashnik  which is widely used as a textbook in many computer science departments.

Another complete proof can be found in the book written by Egorychev which is directly using the inversion rule of residue, see page 49 in the book "Integral representation and the computation of combinatorial sums". If you are reading that book, keep in mind that the index under $\sum$ can be extended to infinity because the contour integral has no pole when $k&gt;n$.
This is a nice example to apply the inversion rule of formal power series stated as rule 4 in G.P. Egorychevs Integral Representations  and the Computation of Combinatorial Sums section 1.2.2.

I think it's worthwhile to present a complete proof. Here we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a formal power series. We show


  The following is valid for $m,n\geq 1$ and $a\in\mathbb{R}$ appropriate
  \begin{align*}
\sum_{{i+j=m}\atop{i,j\geq 0}}&amp;\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j}
=\frac{n}{am+n}\binom{am+n}{m}
\end{align*}


$$ $$


  We obtain
  \begin{align*}
\sum_{{i+j=m}\atop{i,j\geq 0}}&amp;\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{aj+1}\binom{aj+1}{j}\\
&amp;=\sum_{i=0}^m\frac{n-1}{ai+n-1}\binom{ai+n-1}{i}\frac{1}{am-ai+1}\binom{am-ai+1}{m-i}\\
&amp;=\sum_{i=0}^\infty\left\{\binom{ai+n-1}{i}-a\binom{ai+n-2}{i-1}\right\}\\
&amp;\qquad\qquad\cdot\left\{\binom{am-ai+1}{m-i}-a\binom{am-ai}{m-i-1}\right\}\tag{1}\\
&amp;=\sum_{i=0}^{\infty}\left([z^i](1+z)^{ai+n-1}-a[z^{i-1}](1+z)^{ai+n-2}\right)\\
&amp;\qquad\qquad\cdot\left([w^{m-i}](1+w)^{a(m-i)+1}-a[w^{m-i-1}](1+w)^{a(m-i)}\right)\\
&amp;=[w^m](1+w)^{am}(1+w-aw)\\
&amp;\qquad\qquad\cdot\sum_{i=0}^{\infty}\left(\frac{w}{(1+w)^a}\right)^i[z^i](1+z)^{n-2}(1+z-az)(1+z)^{ai}\tag{2}\\
&amp;=[w^m](1+w)^{am}(1+w-aw)\left.\left((1+z)^{n-1}\right)\right|_{z=w}\tag{3}\\
&amp;=[w^m](1+w)^{am+n}-a[w^{m-1}](1+w)^{am+n-1}\\
&amp;=\binom{am+n}{m}-a\binom{am+n-1}{m-1}\\
&amp;=\frac{n}{am+n}\binom{am+n}{m}
\end{align*}
  and the claim follows.


Comment:


In (1) we use the identity $$\frac{q}{pk+q}\binom{pk+q}{k}=\binom{pk+q}{k}-p\binom{pk+q-1}{k-1}$$ We also set the upper limit of the sum to $\infty$ without changing anything since we are only adding zero.
In (2) we rearrange the sum, use the linearity of the coefficient of operator and the rule $[z^{n-k}]A(z)=[z^n]z^kA(z)$.
In (3) we use the inversion rule


\begin{align*}
\sum_{i=0}^\infty w^i [z^i]A(z)f^i(z)=\left.\left(A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}\right)\right|_{z=g(w)}
\end{align*}
with $g(w)$ the inverse of $w=\frac{z}{g(z)}$. We get from (2)
\begin{align*}
A(z)&amp;=(1+z)^{n-2}(1+z-az)\\
f(z)&amp;=(1+z)^a
\end{align*}
and obtain
\begin{align*}
A(z)\frac{f(z)}{f(z)-zf^{\prime}(z)}&amp;=(1+z)^{n-2}(1+z-az)\frac{(1+z)^a}{(1+z)^a-az(a+z)^{a-1}}\\
&amp;=(1+z)^{n-1}
\end{align*}
Since $w=\frac{z}{f(z)}=\frac{z}{(1+z)^a}$ we apply in (3) the substitution $z=w$.
Follow the same reasoning when proving this identity, 
$$\displaystyle \sum_{i=0}^m\frac{q }{pi+q}\binom{pi+q}{i} \binom{pm-pi}{m-i} = \binom{mp+q}{m}$$
You can just read the book yourself. Too difficult to post the book here!

it can be proved using the inversion rule of residue, see page 49 Integral representation and the computation of combinatorial sums