if $p_L=\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|$ is a Cauchy sequence then $\sum_{i\text{ or }j>L}|a_i|\cdot |b_j|<\varepsilon$

Question

I don't understand how inequality $(1)$ is derived in the following proof (where $\{a_n\}$ and $\{b_n\}$ are assumed to converge absolutely).

PROOF
Notice that the sequence $$p_L=\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|$$ converges, since $\{a_n\}$ and $\{b_n\}$ are absolutely convergent, and since the limt of a product is the product of the limits. So $\{p_L\}$ is a Cauchy sequence, which means that for any $\varepsilon>0$, if $L$ and $L'$ are large enough, then $$\left| \sum_{i=1}^{L'}|a_i|\cdot \sum_{j=1}^{L'} |b_j|-\sum_{i=1}^{L}|a_i|\cdot \sum_{j=1}^{L} |b_j|\right|<\frac{\varepsilon}{2}.$$ It follow that $$\sum_{i\text{ or }j>L}|a_i|\cdot |b_j|\leq\frac{\varepsilon}{2}<\varepsilon. \tag{1}$$ $[\dots]$

(Calculus 3rd edition, Michael Spivak. Theorem 23-9, PROOF.)

I must be missing something very simple but can't seem to understand how $(1)$ follows from what came before it.

Answer 1

WLOG assume $L' > L$. Writing as a double sum and splitting into four blocks,

$$\sum_{i=1}^{L'}|a_i| \sum_{j=1}^{L'}|b_j|= \sum_{i=1}^{L'}\sum_{j=1}^{L'}|a_i||b_j| \\= \sum_{i=1}^{L}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=1}^{L}\sum_{j=L+1}^{L'}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=L+1}^{L'}|a_i||b_j|, $$

Hence,

$$\sum_{i=1}^{L'}|a_i| \sum_{j=1}^{L'}|b_j|- \sum_{i=1}^{L}|a_i| \sum_{j=1}^{L}|b_j|\\ = \underbrace{\sum_{i=1}^{L}\sum_{j=L+1}^{L'}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=1}^{L}|a_i||b_j|+ \sum_{i=L+1}^{L'}\sum_{j=L+1}^{L'}|a_i||b_j|}_{i \text{ or }j > L}$$