Product of Sums

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

Then:
 * $\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{\tuple {i, j} \mathop \in A \times B} a_i b_j$

Proof
We have that both series are absolutely convergent.

Thus by Manipulation of Absolutely Convergent Series, it is permitted to expand the product as:


 * $\ds \paren {\sum_{i \mathop \in A} a_i} \paren {\sum_{j \mathop \in B} b_j} = \sum_{i \mathop \in A} \paren {a_i \sum_{j \mathop \in B} b_j}$

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

Hence the result.