Product of Sums

Theorem
If $$\sum_{n\in A} a_n \ $$ and $$\sum_{n\in B} b_n \ $$ are absolutely convergent, then

$$\left({ \sum_{i\in A} a_i }\right) \left({ \sum_{j\in B} b_j }\right) = \sum_{(i,j)\in A \times B} a_i b_j \ $$

Proof
Since both series are absolutely convergent, it is permitted to expand the product as

$$\left({ \sum_{i\in A} a_i }\right) \left({ \sum_{j\in B} b_j }\right) = \sum_{i\in A} \left({ a_i  \sum_{j\in B} b_j }\right) \ $$

but since $$a_i \ $$ is a constant, it may be brought into the sum

$$ = \sum_{i\in A} \sum_{j\in B} a_i b_j  \ $$

which is precisely the theorem.

Corollary
If $$\sum_{i \in X}^\infty a_{ij} \ $$ is absolutely convergent $$\forall j\in Y \ $$, then

$$\prod_{j \in Y}^\infty \left({ \sum_{i \in X}^\infty a_{ij} }\right) = \sum \phi \ $$

where $$\phi \ $$ is every possible product of the form $$a_{xj_1}a_{yj_2}a_{zj_2} \dots \ $$, where $$j_1, j_2, j_3 \dots \ $$ are an enumeration of the elements of $$ Y \ $$.

Proof
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,

$$\prod_{j=1,2} \left({ \sum_{i \in \N} a_{ij} }\right) = \sum_{x,y\in \N} a_{x1}a_{y2} \ $$

and continuing in this vein,

$$\prod_{j=1,2,3} \left({ \sum_{i \in \N} a_{ij} }\right) = \left({ \sum_{x,y\in \N} a_{x1}a_{y2} }\right) \left({ \sum_{z\in \N} a_{z3} }\right) = \sum_{x,y,z\in \N} a_{x1}a_{y2}a_{z3} \ $$

For an inductive proof of this concept for finite $$n \ $$, we assume that for some $$n \in \N \ $$,

$$\prod_{j=1}^n \left({ \sum_{i \in \N} a_{ij} }\right) = \sum_{u,v,\dots,x,y \in \N} a_{u1}a_{v2}\dots a_{x(n-1)}a_{yn} \ $$

Then

$$\prod_{j=1}^{n+1} \left({ \sum_{i \in \N} a_{ij} }\right) = \left({ \sum_{u,v,\dots,x,y \in \N} a_{u1}a_{v2}\dots a_{x(n-1)}a_{yn} }\right) \left({ \sum_{z\in \N} a_{zn} }\right) \ $$

which by the main theorem is simply

$$= \sum_{u,v,\dots,x,y,z \in \N} a_{u1}a_{v2}\dots a_{x(n-1)}a_{yn}a_{z(n+1)} \ $$

completing the induction for finite $$n \ $$