Find generating function for $\sum_{n\ge 0} \binom{m+n}{n}x^n$, $m \in\mathbb{Z}$ I’ve tried applying Vandermonde’s identity, but got stuck. Any help would be appreciated!
 A: For non-negative integers $m,n$,
$$
\sum_{n\ge 0} \binom{m+n}{n}x^n
 = \frac{1}{(1-x)^{m+1}}
$$
It is a classical result, referenced e.g. in the Wiki table of generating functions.
A: Expanding on the previous answer, I'll prove this in a few different ways.
Taylor Series
Using the Taylor series of $(1+x)^s$ with $s$ any real number, we have
$$
(1+x)^s = \sum_{n=0}^\infty \frac{(s)_n}{n!}x^n
$$
where $(s)_n = s(s-1)\cdots (s-n+1)$ with convention $(s)_0 = 1$. This is called the falling factorial. If $s=-(m+1)$ with $m$ an integer, then
$$
(s)_n = (-1)^n (m+1)(m+2)\cdots(m+n) = (-1)^n \frac{(m+n)!}{m!}
$$
Therefore, $(1-x)^{-(m+1)}=\sum_{n=0}^\infty \binom{n+m}{n}x^n$.
Vandermonde Identity Using the Vandermonde identity
$$
\sum_{n=0}^\infty \binom{m+n}{n} x^n = \sum_{n=0}^\infty x^n\sum_{k=0}^n \binom{m}{k}\binom{n}{n-k} = \sum_{k=0}^\infty\binom{m}{k}\sum_{n=k}^\infty\binom{n}{n-k} x^n
$$
where we have switched the order of the sums in the above. Further manipulation yields
$$
=\sum_{k=0}^m\binom{m}{k}x^k\sum_{n=0}^\infty\binom{n+k}{n} x^n
$$
Defining $G_k(x)=\sum_{n=0}^\infty \binom{k+n}{n}x^n$ we therefore have
$$
G_m(x) = \sum_{k=0}^m \binom{m}{k}G_k(x) x^k
$$
Since $G_0(x) = (1-x)^{-1}$ is the gemoetric series, by induction on $m$, we have
$$
(1-x^m)G_m(x) = \frac{1}{1-x}\left[\left(1+\frac{x}{1-x}\right)^m - \left(\frac{x}{1-x}\right)^m\right]
$$
Showing that $G_m(x) = (1-x)^{-(m+1)}$.
Third Way Again let $G_m(x)$ be as before. Note that
$$
\frac{G_{m}(x)}{1-x} = \sum_{s=0}^\infty \sum_{t=0}^\infty \binom{m+s}{s}x^{s+t} = \sum_{q=0}^\infty x^{q}\left(\sum_{s=0}^q
\binom{m+s}{s}\right)
$$
The quantity in the parenthesis is $\binom{m+q+1}{q}$. Therefore $G_{m}(x)/(1-x) = G_{m+1}(x)$. Therefore, $G_m(x) = (1-x)^{-(m+1)}$.
