TITLE: How to find convergence region of $\sum_{n\geqslant 0, m \geqslant 0} x^n y^m \binom{n+m}{n}^2$
QUESTION [7 upvotes]: The following two series are special cases of Appell $F_3$ and $F_4$, namely:
$$
  \mathcal{S}_1 = \sum_{n \geqslant 0, m \geqslant 0} \frac{x^n y^m}{\binom{n+m}{n}}
$$
and
$$
  \mathcal{S}_2 = \sum_{n \geqslant 0, m \geqslant 0} \binom{n+m}{n}^2 x^n y^m
$$
How would one establish that $\mathcal{S}_1$ converges for $\{ (x,y)\colon -1<x<1, -1<y<1 \}$, and $\mathcal{S}_2$ converges for $\{ (x,y) \colon \sqrt{|x|} + \sqrt{|y|} < 1\}$.

REPLY [2 votes]: Here I would like to give direct derivations by bounding series $\mathcal{S}_1$ and $\mathcal{S}_2$ with simple series admitting closed forms.
The convergence of $\mathcal{S}_2$ can be established using $(a^2+b^2) \leqslant (|a|+|b|)^2$ as follows:
$$
   \left| \mathcal{S}_2 \right| \leqslant \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n+m}{n}^2 |x|^n |y|^m \leqslant \left( \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n+m}{n} |x|^{n/2} |y|^{m/2} \right)^2 = \left( \frac{1}{1-\sqrt{|x|} - \sqrt{|y|}}\right)^2
$$
The convergence of $\mathcal{S}_1$ follows rather simply from $\binom{n+m}{n} \geqslant 1$:
$$
   \left| \mathcal{S}_1 \right| \leqslant \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n+m}{n}^{-1} |x|^n |y|^m \leqslant \sum_{n=0}^\infty \sum_{m=0}^\infty |x|^n |y|^m \leqslant  \frac{1}{1-|x|} \cdot \frac{1}{1-|y|}
$$