Product of Sums

Theorem
Let $\displaystyle \sum_{n \mathop \in A} a_n$ and $\displaystyle \sum_{n \mathop \in B} b_n$ be absolutely convergent

Then:
 * $\displaystyle \left({ \sum_{i \mathop \in A} a_i }\right) \left({ \sum_{j \mathop \in B} b_j }\right) = \sum_{\left({i, j}\right) \mathop \in A \times B} a_i b_j$.

Corollary
Let $\displaystyle \sum_{i \mathop \in X} a_{ij}$ be absolutely convergent for all $j \in Y$.

Then:
 * $\displaystyle \prod_{j \mathop \in Y} \left({\sum_{i \mathop \in X} a_{ij}}\right) = \sum_{f : Y \to X} \left( \prod_{j \mathop \in Y} a_{f \left({j}\right) j} \right)$

where $f$ runs over all functions from $Y$ to $X$.

Proof
Since both series are absolutely convergent, it is permitted to expand the product as:
 * $\displaystyle \left({\sum_{i \mathop \in A} a_i }\right) \left({\sum_{j \mathop \in B} b_j }\right) = \sum_{i \mathop \in A} \left({a_i \sum_{j \mathop \in B} b_j}\right)$.

But since $a_i$ is a constant, it may be brought into the sum, so we obtain:
 * $\displaystyle \sum_{i \mathop \in A} \sum_{j \mathop \in B} a_i b_j$

Hence the result

Proof of Corollary
We will prove the case $X = Y = \N$ to avoid the notational inconvenience of enumerating the elements of $Y$ as $j_1, j_2, j_3 \dots$. The general case where $X, Y$ are arbitrary sets has the same proof, but with more indices and notational distractions.

Consider that by the main theorem:
 * $\displaystyle \prod_{j \mathop = 1, 2} \left({\sum_{i \mathop \in \N} a_{ij} }\right) = \sum_{x, y \mathop \in \N} a_{x_1}a_{y_2}$

and continuing in this vein:
 * $\displaystyle \prod_{j \mathop = 1, 2, 3} \left({\sum_{i \mathop \in \N} a_{ij} }\right) = \left({\sum_{x, y \mathop \in \N} a_{x_1} a_{y_2} }\right) \left({\sum_{z \mathop \in \N} a_{z_3} }\right) = \sum_{x, y, z \mathop \in \N} a_{x_1} a_{y_2} a_{z_3}$

For an inductive proof of this concept for finite $n$, we assume that for some $n \in \N$:
 * $\displaystyle \prod_{j \mathop = 1}^n \left({ \sum_{i \mathop \in \N} a_{ij} }\right) = \sum_{u, v, \dots, x, y \mathop \in \N} a_{u_1} a_{v_2}\dots a_{x_{(n-1)}} a_{y_n}$

Then:
 * $\displaystyle \prod_{j \mathop = 1}^{n+1} \left({ \sum_{i \mathop \in \N} a_{ij} }\right) = \left({ \sum_{u, v, \dots, x, y \mathop \in \N} a_{u_1} a_{v_2}\dots a_{x_{(n-1)}} a_{y_n} }\right) \left({\sum_{z \mathop \in \N} a_{z_n} }\right)$

which by Products of Sums is simply:
 * $\displaystyle \sum_{u, v, \ldots, x, y, z \mathop \in \N} a_{u_1} a_{v_2} \ldots a_{x_{(n-1}} a_{y_n} a_{z_{(n+1)}}$

completing the induction for finite $n$.