Fubini's theorem for series

Remarks

For part 1, every finite subset of $\N \times \N$ lies in a rectangle, and every iterated partial sum $\sum_{m \le M} \sum_{n \le N}$ is a finite sum; comparing suprema gives the equality. For part 2, split the real and imaginary parts into positive and negative parts, each bounded by $\abs{a_{m,n}}$, and apply part 1 to each. Summability of a complex family over $\N \times \N$ is equivalent to $\sum_{m} \sum_{n} \abs{a_{m,n}} < \infty$, by part 1. The statement is the case of the counting measure in Tonelli's theorem and Fubini's theorem. Example: for $\abs{x} < 1$ and $\abs{y} < 1$, the family $x^m y^n$ is summable, and its sum is $\frac{1}{1 - x} \cdot \frac{1}{1 - y}$. Without summability the conclusion fails: interchange of summation order gives a double sequence whose two iterated sums differ.

Going deeper

For a student

The theorem is the discrete form of Fubini's theorem, and it is used constantly with power series and Dirichlet series. Typical uses are the Cauchy product of absolutely convergent series, the rearrangement of a double power series, and the exchange of two sums in number theory. The nonnegative case needs no hypothesis at all, which makes it the usual first step: one checks $\sum_m \sum_n \abs{a_{m,n}} < \infty$ and then applies part 2.