Taylor double series For the domain $\{(x+y): -1 < x + y < 1\}$ consider the function
$$f(x,y) = \frac{1}{1 - (x+y)}$$ 
This can be expanded as a convergent geometric series
$$f(x,y) = \sum_{n=0}^\infty(x+y)^n$$
I also see that $(x+y)^n$ can be further expanded as $\sum_{i = 0}^n \frac{n!}{i!(n-i)!}x^i y^{n-i}$.
But how does this relate to the Taylor expansion around (0,0):
$$f(x,y) = \sum_{n=0}^\infty \sum_{m= 0}^\infty \frac{\partial^{n+m}f}{\partial^n x \partial^m y}(0,0) \frac{x^ny^m}{n!m!}$$
Are the single and double series the same thing with the same convergence domains?
 A: The double Taylor series for $f(x,y) = (1 - x - y)^{-1}$ around the origin is 
$$\tag{1}\sum_{n=0}^\infty \sum_{m=0}^\infty \left.\frac{\partial^{n+m}f}{\partial^nx \partial^m y}\right|_{(0,0)}\frac{x^ny^m}{n!m!} = \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{(n+m)!}{n!m!} x^ny^m.$$
On the other hand, using the binomial theorem we have the expansion
$$\tag{2}\sum_{n=0}^\infty (x+y)^n = \sum_{n=0}^\infty \sum_{k=0}^n \frac{n!}{k!(n-k)!}x^ky^{m-k}.$$
Series (1) and (2) are related in that partial sums of (2) coincide with partial sums of (1) taken in a specific order -- summing along diagonals of the matrix $[a_{nm}]$ of general terms.
If the double series (1) is absolutely convergent, then (1) and (2) will converge to the same value.
However, domains of convergence are not identical. For (2) we have absolute convergence for $|x + y| < 1$, as you mentioned, and divergence for $|x+y| \geqslant 1$. This is simply a property of the geometric series.
Determining the convergence domain for (1) is more involved. For positive integers $N$ and $M$, the partial sums satisfy
$$\sum_{n=0}^N \sum_{m= 0}^M \frac{(n+m)!}{n!m!}|x|^n|y|^m < \sum_{n =0}^{N+M}\sum_{k=0}^n \frac{n!}{k!(n-k)!} |x)^k |y|^{n-k} = \sum_{n = 0}^{N+M} (|x| + |y|)^n$$
which should be apparent because every term on the LHS appears on the RHS along with additional terms.
Thus, (1) is absolutely convergent for $|x| + |y| < 1$, and using a similarly constructed lower bound not absolutely convergent if $|x| + |y| \geqslant 1$. The case where $|x + y| < 1 < |x| + |y|$ can arise where the series are not both absolutely convergent. 
