# 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.

-

2
There are two hints: (1) The infinite summation of a non-negative sequence is equal to the supremum of the set of partial sums: $$\sum_{i=1}^{\infty} |a_i| = \sup_{n\in\mathbb{N}} \sum_{i=1}^{n} |a_i|,$$ and (2) supremum can be distributed: $$\sup_{(x,y)\in A\times B}f(x, y) = \sup_{x\in A}\sup_{y\in B} f(x,y).$$ – sos440 Aug 15 '12 at 5:16

## 2 Answers

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 $ML$).

-

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 37, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9593819379806519, "perplexity_flag": "head"}