Products of Sums

Theorem

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

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

Corollary

If $\displaystyle \sum_{i \in X} a_{ij}$ is absolutely convergent $\forall j\in Y$, then:

$\displaystyle \prod_{j \in Y} \left( \sum_{i \in X} a_{ij} \right) = \sum_{\sigma : Y \to Y} \left( \prod_{(i,j) \in X \times Y} a_{i \sigma(j)} \right)$

where $\sigma$ runs over all bijections (permutations in the finite case) from $Y$ to $Y$.

Proof

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

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

But since $a_i$ is a constant, it may be brought into the sum, so we obtain:

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

which is precisely the theorem.

$\blacksquare$

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:

$\displaystyle \prod_{j=1,2} \left({ \sum_{i \in \N} a_{ij} }\right) = \sum_{x,y\in \N} a_{x_1}a_{y_2}$

and continuing in this vein:

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

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

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

Then:

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

which by the main theorem is simply:

$\displaystyle \sum_{u,v,\dots,x,y,z \in \N} a_{u_1}a_{v_2}\dots a_{x_{(n-1)}}a_{y_n}a_{z_{(n+1)}}$

completing the induction for finite $n$.