Proof of product summation of binomial coefficients when I try to proof the sum of two independent negative binomial distribution to be negative binomial, I end up with how to proof the following identity. I try the induction but after I rearrange the terms and use C(n, m)=C(n-1,m-1)+C(n-1,m), it still not completely finish the proof. could you please give me some hint? Thank you.
Proof

$$\sum_{j=0}^k\binom{j+r-1}{j}\cdot\binom{k-j+s-1}{k-j} = \binom{k+r+s-1}{k}$$

Source.
 A: We use the coefficient of operator $[x^j]$ to denote the coefficient of $x^{j}$ in a polynomial or series $A(x)=\sum_{k=0}^{\infty}a_kx^k$. We can write e.g.
\begin{align*}
  \binom{k}{j}=[x^j](1+x)^k
  \end{align*}

We obtain
  \begin{align*}
\sum_{j=0}^k&\binom{j+r-1}{j}\binom{k-j+s-1}{k-j}\\
&=\sum_{j=0}^{\infty}\binom{-r}{j}(-1)^j\binom{-s}{k-j}(-1)^{k-j}\tag{1}\\
&=(-1)^k\sum_{j=0}^\infty[x^j](1+x)^{-r}[y^{k-j}](1+y)^{-s}\tag{2}\\
&=(-1)^k[y^k](1+y)^{-s}\sum_{j=0}^{\infty}y^j[x^j](1+x)^{-r}\tag{3}\\
&=(-1)^k[y^k](1+y)^{-(r+s)}\tag{4}\\
&=\binom{-(r+s)}{k}(-1)^k\\
&= \binom{r+s+k-1}{k}
\end{align*}

Comment:


*

*In (1) we use $\binom{-n}{k}(-1)^k=\binom{n+k-1}{k}$ twice and set the upper limit of the sum to $\infty$ without changing anything, since we add only zero.

*In (2) we use the coefficient of operator twice 

*In (3) we rearrange the sum by using the linearity of the coefficient of operator and $[x^{n+k}]A(x)=[x^n]x^{-k}A(x)$.

*In (4) we apply the substitution rule
\begin{align*}
  A(x)=\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}x^k[y^k]A(y)
  \end{align*}
with $a_k=[y^k]A(y)$.
