A question on iterated sums We are to prove that the set
$$\Bigg\{ \sum^m_{i=1}\sum^n_{j=1}|a_ib_j| :m,n\in \mathbb N \Bigg\}$$
is bounded. Also, We are to use this to show that the iterated sum $\sum^\infty_{i=1} \sum^\infty_{j=1}|a_ib_j|$ converges.
I could prove that the set is bounded, however, I can't solve the second part.
I become very confused when I have to deal with iterated sums. Could anyone please help me with this problem.
 A: In order to show that $\sum_{i=1}^\infty\sum_{j=1}^\infty|a_ib_j|$ converges, you must show that there is some real number $a$ such that for every $\epsilon>0$ there is some $n_0\in\Bbb Z^+$ such that $$\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$$ whenever $m,n\ge n_0$.
For $m,n\in\Bbb Z^+$ let $$s_{mn}=\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\;.$$ You know that the set $S=\{s_{mn}:m,n\in\Bbb Z^+\}$ is bounded. You also know that $s_{k\ell}\le s_{mn}$ whenever $k\le m$ and $\ell\le n$, since the terms $|a_ib_j|$ are non-negative. For $n\in\Bbb Z^+$ let $d_n=s_{nn}$; then $s_{k\ell}\le d_n$ whenever $k,\ell\le n$. In other words, the sums $d_n$ are cofinal in $S$: for each $s\in S$ there is an $n\in\Bbb Z^+$ such that $s\le d_n$.
Consider the sequence $\langle d_n:n\in\Bbb Z^+\rangle$: it’s bounded and non-decreasing, so it converges to some limit $a$; I’ll leave to you the easy verification that $s\le a$ for all $s\in S$. Fix $\epsilon>0$. There is some $n_0\in\Bbb Z^+$ such that $|a-d_n|<\epsilon$ whenever $n\ge n_0$. Suppose that $m,n\ge n_0$; then $a-\epsilon<d_{n_0}\le s_{mn}\le a$, so $$\left|a-\sum_{i=1}^m\sum_{j=1}^n|a_ib_j|\right|<\epsilon$$ whenever $m,n\ge n_0$, exactly as we wanted.
A: Although by first proving that the sum is bounded, it can be proved that the sum converges. But, there is a way to directly prove that it converges.
For that, let 
$$\sum^\infty_{i=1}|a_i|=L~~~~~\text{and}~~~~~\sum^\infty_{j=1}|b_j|=M$$
For each fixed $i \in \mathbb N$, the Algebraic Limit Theorem will allow us to write $\sum^\infty_{j=1}|a_ib_j|$=$|a_i|\sum^\infty_{j=1}|b_j|$. 
If one continues the process, we will see that:
$$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|=\sum^\infty_{i=1}|a_i|\sum^\infty_{j=1}|b_j|=ML$$
Hence,$\sum^\infty_{i=1}\sum^\infty_{j=1}|a_ib_j|$ converges  (to $ML$).
