# Doubt about a proof of “Every absolutely convergent series is commutative convergent”

In the prove of my book for the theorem which says that every absolutely convergent series is commutative convergent, where $$\Sigma a_n$$ is absolutely convergent, $$s_n = a_1 + ... + a_n$$, $$\Sigma b_n$$ is a series which elements are the $$\Sigma a_n$$ elements switched and $$t_n = b_1 + ... + b_n$$, it is written:

"Being $$\sigma:\Bbb{N}\rightarrow\Bbb{N}$$ a bijection and defining $$b_n = a_{\sigma(n)}$$. For each $$n\in\Bbb{N}$$, we will call by $$m$$ the biggest number of $$\sigma(1),\sigma(2),...,\sigma(n)$$. So $${\sigma(1),...,\sigma(n)}\subset[1,m]$$. Then we have that $$t_n=\sum_{i = 1}^na_{\sigma(i)}\le\sum_{j = 1}^ma_j=s_m$$. Thus for each $$n\in \Bbb{N}$$ there is a $$m\in \Bbb{N}$$ such that $$t_n \ge s_m$$ and, by an analogous way, for each $$m\in \Bbb{N}$$ there is a $$n\in \Bbb{N}$$, such that $$s_m \ge t_n$$. So $$\lim s_n=\lim t_n$$"

The part that I didn't understand:

"for each $$n\in \Bbb{N}$$ there is a $$m\in \Bbb{N}$$ such that $$t_n \ge s_m$$ and, by an analogous way, for each $$m\in \Bbb{N}$$ there is a $$n\in \Bbb{N}$$, such that $$s_m \ge t_n$$. So $$\lim s_n=\lim t_n$$"

(Sorry if something is grammatically wrong, the book isn't in English and I am trying to translate it.)

Suppose that $$\lim s_n = L \neq M = \lim t_n$$. You can choose $$\epsilon > 0$$ such that $$(L-\epsilon,L+\epsilon) \cap (M-\epsilon,M+\epsilon) = \emptyset$$ and $$s_n \in (L-\epsilon,L+\epsilon);\quad t_n \in (M-\epsilon,M+\epsilon)$$ as $$n \geq n_0 \in \Bbb{N}$$ and only finite elements of $$(s_n)$$ and $$(t_n)$$ may not be in these intervals. Also, you can write $$t_{k_1} \geq s_{k_2} \geq t_{k_3} \geq s_{k_4} \geq t_{k_5} \geq s_{k_6} \geq \cdots\tag{1}$$ but $$s_n \geq t_n; \forall n>n_0$$ or $$s_n \leq t_n; \forall n>n_0$$. It means that $$(1)$$ has only finite elements, a contradiction.

• Sorry but can you explain me why (1) would have only finite elements? – Rebeca Lie Yatsuzuka Silva Jan 17 '20 at 17:25

What you wrote is not enough. From $$\forall n \exists m,\quad t_n \ge s_m \\ \forall m \exists n,\quad t_n \le s_n$$ is not enough to show the limits are the same. (Example below.)

What we do need is something like $$\forall n \exists m \ge n,\quad t_n \ge s_m \\ \forall m \exists n \ge m,\quad t_n \le s_n$$ Presumably that is what is actually proved in your book (whose identity is secret).

Example
$$t_1 = s_1 = 10$$; $$t_2 = s_2 = 0$$, for $$n \ge 3, t_n = 2, s_n = 5$$.

Then: for all $$n$$ there exists $$m$$ (namely $$m=2$$) so that $$t_n \ge s_m$$.
And: for all $$m$$ there exists $$n$$ (namely $$n=2$$) so that $$t_n \le s_m$$.

I assume that the book that you were studying was "Curso de Análise Vol.1" by Elon Lages Lima, since I had the same problem of not understanding the middle steps of the proof.

It is important to note that the proof given is only for $$a_n \geq 0$$, and that the general case for $$\sum a_n$$ absolutely convergent follows by the separation of $$a_n$$ by positive terms and negative terms such that $$\sum a_n = \sum p_n - \sum q_n$$.

Let $$s_n = a_1 + ... + a_n$$ and $$t_n = b_1 + ... + b_n$$. Since $$a_n \geq 0$$ for all $$n$$, the sequences $$\{s_n\}$$ and $$\{t_n\}$$ are non-decreasing. Furthermore, since $$\{s_n\}$$ is convergent, we have that:

$$S = \sum^\infty a_n = \lim s_n = \sup\{s_n: n\in \mathbb{N}\}$$

then, it suffices to prove that:

i) $$\{t_n\}$$ is bounded above, which implies that $$T = \sum^\infty b_n = \lim t_n = \sup\{t_n: n\in \mathbb{N}\}$$

ii) $$\sup \{s_n: n \in \mathbb{N}\} = \sup \{t_n: n\in \mathbb{N}\}$$

Now for every $$n\in \mathbb{N}$$, let $$m = \max\{\sigma(1), ...., \sigma(n)\}$$. It follows that:

$$t_n = \sum^n_{i = 1} a_{\sigma(i)} \leq \sum_{i=1}^m a_k = s_m \leq S$$

thus, $$\forall n \in \mathbb{N}\exists{m}\in \mathbb{N}: t_n \leq s_m \leq S$$, so $$\{t_n\}$$ is bounded above, and then $$\sum b_n$$ is convergent with $$\sum b_n = \sup \{t_n: n\in \mathbb{N}\}$$.

The last inequality also shows that $$\sup \{t_n\} \leq S$$. By an analogous way (using $$\sigma^{-1}$$) we can show that $$\forall m \in \mathbb{N}\exists n \in \mathbb{N}: s_m \leq t_n \leq T$$, which implies that $$S \leq T$$. Therefore, $$S = T$$, that is $$\lim t_n = \sum b_n = \sum a_n = \lim s_n$$.