# How to find convergence region of $\sum_{n\geqslant 0, m \geqslant 0} x^n y^m \binom{n+m}{n}^2$

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 $\{ (x,y) \colon \sqrt{|x|} + \sqrt{|y|} < 1\}$.

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You can use the ratio test for double series.

Here is a good reference, just look under the ratio test theorem:

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+1 Thank you for the valuable reference. The ratio test, being really a divergence (rather than convergence) test, is not able to establish the region of convergence of $\mathcal{S}_2$. One has to bound the double series with a computable one. – Sasha Jul 17 '12 at 15:58
By the way, I did not pay attention to the second series. – Mhenni Benghorbal Jul 17 '12 at 17:54
But you can use the ratio test to prove $S_{1}$. On the other hand, you used the comparison test theorem which is included in the paper. – Mhenni Benghorbal Jul 18 '12 at 1:45

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 $\binom{n+m}{n} \geqslant 1$: $$\left| \mathcal{S}_1 \right| \leqslant \sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n+m}{n}^{-1} |x|^n |y|^m \leqslant \sum_{n=0}^\infty \sum_{m=0}^\infty |x|^n |y|^m \leqslant \frac{1}{1-|x|} \cdot \frac{1}{1-|y|}$$

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