I am reading "Calculus 3rd Edition" by Michael Spivak.
What does $\sum_{i \text{ or } j > L} |a_i| \cdot |b_j|$ mean precisely?
What is the definition of $\sum_{i \text{ or } j > L} |a_i| \cdot |b_j|$?
I guess $$\sum_{i \text{ or } j > L} |a_i| \cdot |b_j| := \lim_{L \to \infty} p_L – \sum_{i=1}^{L} |a_i| \cdot |\sum_{j=1}^{L} |b_j|.$$
If we replace $$\sum_{i \text{ or } j > L} |a_i| \cdot |b_j|$$ with $$\lim_{L \to \infty} p_L – \sum_{i=1}^{L} |a_i| \cdot |\sum_{j=1}^{L} |b_j|,$$ the proof works as well.
THEOREM 9
If $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ converge absolutely,and $\{c_n\}$ is any sequence containing the product $a_i b_j$ for each pair $(i, j)$, then $$\sum_{n=1}^{\infty} c_n = \sum_{n=1}^{\infty} a_n \cdot \sum_{n=1}^{\infty} b_n.$$
PROOF
Notice first that the sequence $$p_L = \sum_{n=1}^{L} |a_n| \cdot \sum_{n=1}^{L} |b_n|$$
converges, since $\{a_n\}$ and $\{b_n\}$ are absolutely convergent, and since the limit of a product is the product of the limits. So $\{p_L\}$ is a Cauchy sequence, which means that for any $\epsilon > 0$, if $L$ and $L'$ are large enough, then $$\left |\sum_{i=1}^{L'} |a_i| \cdot \sum_{j=1}^{L'} |b_j| – \sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j|\right | < \frac{\epsilon}{2}\tag{1}$$
It follows that $$\sum_{i \text{ or } j > L} |a_i| \cdot |b_j| \leq \frac{\epsilon}{2} < \epsilon.$$
Now suppose that $N$ is any number so large that the terms $c_n$ for $n \leq N$ include every term $a_i b_j$ for $i, j \leq L$. Then the difference $$\sum_{n=1}^{N} c_n – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j$$
consists of terms $a_i b_j$ with $i > L$ or $j > L$, so
$$|\sum_{n=1}^{N} c_n – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| \leq \sum_{i \text{ or } j > L} |a_i| \cdot |b_j| < \epsilon.$$
But since the limit of a product is the product of the limits, we also have $$|\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| < \epsilon$$ for large enough $L$. Consequently, if we choose $L$, and then $N$, large enough, we will have $$|\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{n=1}^{N} c_n| \leq |\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| + |\sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j – \sum_{n=1}^{N} c_n| < 2 \epsilon,$$ which proves the theorem.
MY PROOF WITHOUT $\sum_{i \text{ or } j > L} |a_i| \cdot |b_j|$
Notice first that the sequence $$p_L = \sum_{n=1}^{L} |a_n| \cdot \sum_{n=1}^{L} |b_n|$$
converges, since $\{a_n\}$ and $\{b_n\}$ are absolutely convergent, and since the limit of a product is the product of the limits. So for any $\epsilon > 0$, if $L$ is large enough, then $$\sum_{i=1}^{\infty} |a_i| \cdot \sum_{j=1}^{\infty} |b_j| – \sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j| < \epsilon.$$
Now suppose that $N$ is any number so large that the terms $c_n$ for $n \leq N$ include every term $a_i b_j$ for $i, j \leq L$. Then the difference $$\sum_{n=1}^{N} c_n – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j$$
consists of terms $a_i b_j$ with $i > L$ or $j > L$, so
$$\sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j| + |\sum_{n=1}^{N} c_n – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| \leq \sum_{i=1}^{M} |a_i| \cdot \sum_{j=1}^{M} |b_j| \leq \sum_{i=1}^{\infty} |a_i| \cdot \sum_{j=1}^{\infty} |b_j|$$ for large enough $M$.
So, $$|\sum_{n=1}^{N} c_n – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| \leq \sum_{i=1}^{\infty} |a_i| \cdot \sum_{j=1}^{\infty} |b_j| – \sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j| < \epsilon.$$
But since the limit of a product is the product of the limits, we also have $$|\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| < \epsilon$$ for large enough $L$. Consequently, if we choose $L$, and then $N$, large enough, and then $M$, large enough, we will have $$|\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{n=1}^{N} c_n| \leq |\sum_{i=1}^{\infty} a_i \cdot \sum_{j=1}^{\infty} b_j – \sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j| + |\sum_{i=1}^{L} a_i \cdot \sum_{j=1}^{L} b_j – \sum_{n=1}^{N} c_n| < 2 \epsilon,$$ which proves the theorem.
Best Answer
Starting with inequality (1)
$$\left |\sum_{i=1}^{L'} |a_i| \cdot \sum_{j=1}^{L'} |b_j| - \sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j|\right | < \frac{\epsilon}{2}\tag{1}$$
and assuming $L'>L$, the terms in $\sum_{i=1}^{L} |a_i| \cdot \sum_{j=1}^{L} |b_j|$ all have $i$ and $j$ smaller than or equal to $L$.
When we remove these terms from $\sum_{i=1}^{L'} |a_i| \cdot \sum_{j=1}^{L'} |b_j|$ we are left with terms in which either $i$ or $j$ are larger than $L$.
These terms are denoted
$$\sum_{i \text{ or } j > L} |a_i| \cdot |b_j|$$