If $\{a_n\}$ and $\{b_n\}$ are absolutely summable, why is the arrangement of terms in $\sum_{n=1}^\infty a_n\cdot\sum_{n=1}^\infty b_n$ irrelevant? My question below is based on a section from Chapter 23, "Infinite Series", of Spivak's Calculus.
Consider the task of multiplying two infinite series. That is, we want to compute
$$\left (\sum\limits_{n=1}^\infty a_n \right )\cdot \left ( \sum\limits_{n=1}^\infty b_n \right )=(a_1+a_2+a_3+...)\cdot (b_1+b_2+b_3+...)$$
All possible products $a_ib_j$ can be arranged into a two-dimensional array. We can arrange the elements of this array in a sequence in an infinite number of ways. Here is an example


Let $\{c_n\}$ be a sequence of this sort, containing each product
$a_ib_j$ just once.
We might naively expect to have
$$\sum\limits_{n=1}^\infty c_n=\sum\limits_{n=1}^\infty
a_n\cdot\sum\limits_{n=1}^\infty b_n$$
but this isn't true, nor is this really surprising since we've said
nothing about the specific arrangement of the terms. The next theorem
shows that the result does hold when the arrangement of terms is
irrelevant
Theorem 9 If $\sum\limits_{n=1}^\infty a_n$ and $\sum\limits_{n=1}^\infty b_n$ converge absolutely, and $\{c_n\}$ is
any sequence containing the products $a_ib_j$ for each pair $(i,j)$,
then
$$\sum\limits_{n=1}^\infty c_n=\sum\limits_{n=1}^\infty
 a_n\cdot\sum\limits_{n=1}^\infty b_n$$

Why is the arrangement of terms irrelevant in the scenario in Theorem 9?
 A: Let's analyze the proof of Theorem 9 for insights.
Consider the sequence $p_L=\sum\limits_{i=1}^L |a_i| \sum\limits_{j=1}^L |b_j|$. This is a sequence of partial sums.
For example,
$$p_1=|a_1||b_1|$$
$$p_2=|a_1||b_1|+|a_1||b_2|+|a_2||b_1|+|a_2||b_2|$$
This sequence converges because the following limit exists
$$\lim\limits_{L\to\infty} p_L=\lim\limits_{L\to\infty} \sum\limits_{i=1}^L |a_i| \cdot \lim\limits_{L\to\infty}\sum\limits_{j=1}^L |b_j|\tag{1}$$
We can use this formula equation because both right-hand side limits exist.
Notice that any rearrangement of either $\{a_n\}$ or $\{b_n\}$ leads to the same limits for these sequences and also for $p_L$.
This follows from the following theorem

Theorem 8 If $\sum\limits_{n=1} a_n$ converges absolutely and $\{b_n\}$ is any rearrangement of $\{a_n\}$ then $\sum\limits_{n=1} b_n$ also converges absolutely and $\sum\limits_{n=1} a_n=\sum\limits_{n=1}b_n$

I am speculating here, but I think this means that by rearranging the individual sequences we can get different orders of terms in the partial sums and in $\lim\limits_{L\to\infty} p_L$ and in each case $p_L$ converges to the same number. I am not totally sure about this, because as $p_L$ is constructed it doesn't seem that we get all the possible orders of terms $a_ib_j$.
If $\{a_n\}$ and $\{b_n\}$ were summable but not absolutely summable, then we could not just rearrange the sequences and still be sure that we've maintained
$$\lim\limits_{L\to\infty} p_L=\lim\limits_{L\to\infty} \sum\limits_{i=1}^L a_i \cdot \lim\limits_{L\to\infty}\sum\limits_{j=1}^L b_j\tag{2}$$
Thus, we cannot be sure that $p_L$ converges, since it depends on the order of the terms in the product of the sequences.
In the case of absolute convergence of $\{a_n\}$ and $\{b_n\}$ we know that whatever order of terms we choose in $p_L$ it converges.
Here is the rest of the proof, just for the record. But the main reasoning is above.
Since $\{p_L\}$ is a Cauchy sequence, then for $L$ and $L'$ large enough with $L'>L$ we have
$$\left | \sum\limits_{i=1}^{L'} |a_i|\sum\limits_{j=1}^{L'} |b_j|- \sum\limits_{i=1}^{L} |a_i|\sum\limits_{j=1}^{L} |b_j|\right |<\frac{\epsilon}{2}<\epsilon$$
The left-hand side means we only keep terms of $\sum\limits_{i=1}^{L'} |a_i|\sum\limits_{j=1}^{L'} |b_j|$ for which $i$ or $j$ is larger than $L$. Thus we have
$$\sum\limits_{i\text{ or } j > L}|a_i||b_j|<\frac{\epsilon}{2}<\epsilon$$
Now, let $\{c_n\}$ be any sequence such that each term is of form $a_ib_j$ for each pair $(i,j)$.
Suppose $N$ is large enough that the terms $c_n$ for $n\leq N$ include every term $a_ib_j$ for $i,j<L$.
Then the difference
$$\sum\limits_{n=1}^{N} c_n-\sum\limits_{i=1}^{L} a_i\cdot \sum\limits_{j=1}^{L} b_j$$
consists of terms $a_ib_j$ with $i>L$ or $j>L$, so
$$\left |\sum\limits_{n=1}^{N} c_n-\sum\limits_{i=1}^{L} a_i\cdot \sum\limits_{j=1}^{L} b_j\right |\leq \sum\limits_{i\text{ or } j > L}|a_i||b_j|<\frac{\epsilon}{2}<\epsilon$$
To put it in more words, $\{c_n\}$ is a sequence of terms $a_ib_j$. If the first $N$ terms contain terms $a_ib_j$ for $i,j<L$, then $\sum\limits_{i\text{ or } j > L}a_ib_j$ represents all the (infinite) other terms of this kind, which have $i$ or $j>L$. The rest of the terms in $\sum\limits_{n=1}^{N} c_n$ will be a subset of these infinite other terms.
Then,
$$\left | \sum\limits_{i=1}^\infty a_i \sum\limits_{j=1}^\infty b_j-\sum\limits_{n=1}^{N}c_n\right |$$
$$=\left | \sum\limits_{i=1}^\infty a_i \sum\limits_{j=1}^\infty b_j-\sum\limits_{i=1}^{L} a_i \sum\limits_{j=1}^{L} b_j+\sum\limits_{i=1}^{L} a_i \sum\limits_{j=1}^{L} b_j-\sum\limits_{n=1}^{N}c_n\right |$$
$$=\left | \sum\limits_{i=1}^\infty a_i \sum\limits_{j=1}^\infty b_j-\sum\limits_{i=1}^{L} a_i \sum\limits_{j=1}^{L} b_j\right |+\left |\sum\limits_{i=1}^{L} a_i \sum\limits_{j=1}^{L} b_j-\sum\limits_{n=1}^{N}c_n\right |$$
$$<2\epsilon$$
