Closed-form expression of a series involving combinatorics

Question

How to find the closed-form expression of the function $$F(x,y)=\sum_{n=0}^\infty\sum_{m=0}^\infty {{m+n}\choose {n}}x^ny^m=1+x+y+x^2+y^2+2xy+...?$$

Answer 1

$$ \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n+m}{n}x^n y^m = \sum_{n=0}^\infty \sum_{m=0}^n \binom{n}{n-m}x^{n-m}y^{m}= \sum_{n=0}^\infty (x+y)^n = \frac{1}{1-x-y} $$ Due to the following fact: let $a_{m,n} n,m\ge 0$ be a doubly indexed sequence s.t. $\sum_{m,n\ge 0}a_{m,n}$ is absolutely convergent. Then $$ \sum_{n,m\ge 0}a_{m,n} = \sum_{k=0}^\infty \sum_{(n,m):n+m=k} a_{n,m} = \sum_{k=0}^\infty \sum_{l=0}^k a_{l,k-l} $$ Now set $a_{m,n} = \binom{n+m}{n} x^n y^m$. Note that $\binom{k}{k-l}=\binom{k}{l}$.

Answer 2

The inner sum is $$\sum_{m=0}^\infty \binom{m+n}{n} y^m=\frac{1}{(1-y)^{n+1}}.$$

So \begin{align} F(x,y)&=\sum_{n=0}^\infty \frac{1}{(1-y)^{n+1}} x^n \\ &= \frac{1}{1-y}\sum_{n=0}^\infty \left(\frac{x}{1-y}\right)^n \\ &= \frac{1}{1-y}\cdot\frac{1}{1-x/(1-y)} \\ &= \frac{1}{1-x-y}. \end{align}