What is the domain of convergence of the double sum $\sum_{i,j \in \mathbb N}\frac{(i+j)!}{i!j!}z^iw^j$? What is the domain of convergence of  the double sum $\sum_{i,j \in \mathbb N}\frac{(i+j)!}{i!j!}z^iw^j$, where $z,w$ are complex numbers? How do you define the partial sum of the double sum really? I cannot find any reference on this matter.
 A: Write this as
$$\sum_{i=0}^\infty \sum_{j=0}^\infty \frac{(i+j)!}{i!j!}z^iw^j- 1 - \sum_{i=0}^\infty z^i-  \sum_{j=0}^\infty w^j,$$
where the single series are easily handled. 
Consider the double series. The partial sum is defined as 
$$S_{mn}(z,w) =\sum_{i=0}^m \sum_{j=0}^n \frac{(i+j)!}{i!j!}z^iw^j.$$
This converges in the strongest sense to $S(z,w)$ if for every $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that if $n,m > N$ we have $\left|S_{mn}(z,w) - S(z,w)\right| < \epsilon. $
There are weaker forms of convergence, such as summing by rows, columns, and diagonals.  In general, convergence in these cases does not guarantee convergence in the strongest sense. 
To sum along diagonals,  we consider convergence of partial sums of the form $S_{m}(z,w)$ where
$$S_m(z,w) = \sum_{p=0}^m \sum_{q=0}^p \frac{p!}{q!(p-q)!}z^qw^{p-q}. $$
Using the binomial theorem we get
$$S_m(z,w) = \sum_{p=0}^m (z+w)^p, $$
which converges as $m \to \infty$ if $|z + w| < 1$.
Addendum
We can show that this double series is absolutely convergent in the strongest sense when $|z| + |w| < 1.$ This is, of course, a stronger condition than is necessary for diagonal convergence since $|z + w| \leqslant |z| + |w|$.
Note that
$$\sum_{i=0}^m \sum_{j=0}^n \frac{(i+j)!}{i!j!}|z|^i|w|^j \leqslant \sum_{p=0}^{m+n} \sum_{q=0}^p \frac{p!}{q!(p-q)!}|z|^q|w|^{p-q} = \sum_{p=0}^{m+n} (|z| + |w|)^p. $$
The double series converges as $m,n \to \infty$ if the geometric series on the RHS converges. This is the case when $|z| + |w| < 1$.
