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.

  • $\begingroup$ Sorry but can you explain me why (1) would have only finite elements? $\endgroup$ – 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).

$t_1 = s_1 = 10$; $t_2 = s_2 = 0$, 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 $\lim t_n = \sum b_n = \sum a_n = \lim s_n$.


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