Generating function of product of binomial coefficients Let $m, n\in \mathbb{N}$ and $|x| < 1$. I look for hints to derive an analytic formula for
$$f_{m,n}(x) = \sum_{k \in \mathbb{N}} {n + k \choose k} {m + k \choose k}  x^{k}. $$
 A: This is Gauss' hypergeometric function
$F(n+1,m+1,1;x)$. You can then apply the huge theory of hypergeometric functions to derive further expressions. For instance, Euler's transformation formula gives the alternative expression
$$\frac 1{(1-x)^{m+n+1}}\,F(-m,-n,1;x)=\frac 1{(1-x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom mk\binom nk x^k$$
for the same series as a finite sum.
A: With proofs abound, I would like to record yet another instance of application due to the Wilf-Zeilberger techniques. The aim is to prove the identity shown above (from GH from MO, Hjalmar Rosengren):
$$\sum_r\binom{m+n+k-r}{k-r}\binom{m}r\binom{n}r\binom{m+k}k^{-1}\binom{n+k}k^{-1}=1.$$
The mechanical process begins with letting (suppressing other variables)
$$F(n,r):=\binom{m+n+k-r}{k-r}\binom{m}r\binom{n}r\binom{m+k}k^{-1}\binom{n+k}k^{-1}$$
and also that
$$G(n,r):=-\frac{F(n,r)\,(m+n+1+k-r)r^2}{(m+n+1)(n+1+k)(n-r+1)}.$$
Next step: verify $F(n+1,r)-F(n,r)=G(n,r+1)-G(n,r)$ which upon summing (both sides) over all integers $k$ results in cancellation on the RHS. Hence, $f(n+1)-f(n)=0$ where $f(n)=\sum_kF(n,r)$. Since a direct computation shows $f(0)=1$, it follows that $f(n)=1$. The proof is now complete. $\,\,\square$
A: Hjalmar Rosengren gave a nice formula for $f_{m,n}(x)$ based on the theory of hypergeometric series. Here I provide a direct elementary proof of the same based on generating series.
Let us introduce the differential operator
$$D_r:=\frac{1}{r!}\frac{\partial^r}{\partial x^r}.$$
Then
$$\sum_{k=0}^\infty\binom{n+k}{k}x^k=D^n\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^{n+1}},$$
hence by the general Leibniz rule
\begin{align*}\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k
&=D^m\left(\frac{x^m}{(1-x)^{n+1}}\right)\\
&=\sum_{k=0}^m D^{m-k}(x^m)D^k\left(\frac{1}{(1-x)^{n+1}}\right)\\
&=\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}\frac{x^k}{(1-x)^{k+n+1}}\\
&=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^m\binom{m}{k}\binom{n+k}{k}x^k(1-x)^{m-k}\\
&=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^m D^n(x^{n+k})\binom{m}{k}(1-x)^{m-k}.
\end{align*}
Here we can interpret $D^n(x^{n+k})$ as the coefficient of $t^n$ in the polynomial $(x+t)^{n+k}\in\mathbb{Z}[x][t]$, hence by the binomial theorem, the $k$-sum on the right-hand side equals
$$[t^n]\sum_{k=0}^m (x+t)^{n+k}\binom{m}{k}(1-x)^{m-k}=[t^n](x+t)^n(1+t)^m.$$
Expanding $(x+t)^n$ and $(1+t)^m$ by the binomial theorem, and then collecting the coefficient of $t^n$ in the product of the resulting two sums, we get that
$$\sum_{k=0}^\infty\binom{m+k}{k}\binom{n+k}{k}x^k=\frac{1}{(1-x)^{m+n+1}}\sum_{k=0}^{\min(m,n)}\binom{m}{k}\binom{n}{k}x^k.$$
