# Theorem 3.55 in Baby Rudin: Every re-arrangement of an absolutely convergent series converges to the same sum in every normed space?

Here's Theorem 3.55 in the book Principles of Mathematical Analysis by Walter Rudin, third edition.

If $\sum a_n$ is a series of complex numbers which converges absolutely, then every rearrangement of $\sum a_n$ converges, and they all converge to the same sum.

And, here's Rudin's proof.

Let $\sum a_n^\prime$ be a rearrangement, with partial sums $s_n^\prime$. Given $\varepsilon > 0$, there exists an integer $N$ such that $m \geq n \geq N$ implies $$\mbox{ (26) } \ \ \ \ \sum_{i=n}^m \vert a_i \vert \leq \varepsilon.$$ Now choose $p$ such that the integers $1, 2, \ldots, N$ are all contained in the set $k_1, k_2, \ldots, k_p$. [Here $\{k_n\}$ is a sequence of positive integers in which every positive integer appears as a term once and only once, and $a_n^\prime = a_{k_n}$ for each $n = 1, 2, 3, \ldots$; moreover, $s_n^\prime = a_1^\prime + \cdots + a_n^\prime$. This is the notation of Definition 3.52 in Rudin. ] Then if $n > p$, the $a_1, \ldots, a_N$ will cancel in the difference $s_n - s_n^\prime$, so that $\vert s_n - s_n^\prime \vert \leq \varepsilon$ by (26). Hence $\{s_n^\prime \}$ converges to the same sum as $\{s_n \}$.

Now here is my reading of Rudin's proof.

As $\sum a_n$ converges absolutely, the series $\sum \vert a_n \vert$ converges, which means that the sequence $\{ \sum_{i =1}^n \vert a_i \vert \}_{n \in \mathbb{N}}$ is convergent and therefore Cauchy. Thus, given a real number $\varepsilon > 0$, we can find a natural number $N$ such that $$\left\vert \sum_{i=1}^n \vert a_i \vert - \sum_{i=1}^m \vert a_i \vert \right\vert < \varepsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that } \ n \geq m > N.$$ That is, $$\sum_{i = m+1}^n \vert a_i \vert < \varepsilon \ \mbox{ for all } \ m, n \in \mathbb{N} \ \mbox{ such that } \ n \geq m > N.$$ Now let $\{k_n \}$ be a sequence of natural numbers in which every natural number appears exactly once, and let $a_n^\prime = a_{k_n}$, and then let $s_n^\prime = \sum_{i=1}^n a_i^\prime$ for each $n = 1, 2, 3, \ldots$. We first need to show that the series $\sum a_n^\prime$ converges. let $p$ be a natural number such that $$\{1, \ldots, N \} \subset \{ k_1, \ldots, k_p \}.$$ Then, for all $n \in \mathbb{N}$ such that $n > p$, we have $$\vert s_n^\prime - s_n \vert = \left\vert \sum_{i \in \{ k_1, \ldots, k_n \} \setminus \{1, \ldots, n \} } a_i \right\vert \leq \sum_{i \in \{ k_1, \ldots, k_n \} \setminus \{1, \ldots, n \} } \vert a_i \vert \leq \sum_{i = N+1}^{\max ( \{ k_1, \ldots, k_n \} \setminus \{1, \ldots, N \} )} \vert a_i \vert < \varepsilon.$$ Now let's suppose that $$\lim_{n \to \infty} s_n = s.$$ Then for $n \in \mathbb{N}$ such that $n > p$, we have $$0 \leq \vert s_n^\prime - s \vert \leq \vert s_n^\prime - s_n \vert + \vert s_n - s \vert < \epsilon + \vert s_n - s \vert \to \epsilon + 0 \ \mbox{ as } \ n \to \infty.$$ So, $$0 \leq \lim_{n \to \infty} \vert s_n^\prime - s \vert \leq \epsilon,$$ provided that the limit exists (i.e. provided that the sequence $\{s_n^\prime \}$ converges). We now show that the sequence $\{s_n^\prime \}$ is Cauchy. For all $m, n \in \mathbb{N}$ such that $n \geq m > p$, we have $$\vert s_n^\prime - s_m^\prime \vert \leq \sum_{i = m+1}^n \vert a_i^\prime \vert \leq \sum_{i = N+1}^{\max( \{ k_{m+1}, \ldots, k_n \} )} \vert a_i \vert < \varepsilon,$$ showing that the sequence $\{s_n^\prime \}$ is indeed Cauchy.

So for all $m, n \in \mathbb{N}$ such that $n \geq m > p$, we have $$\left\vert \vert s_n^\prime - s \vert - \vert s_m^\prime - s \vert \right\vert \leq \left\vert (s_n^\prime - s) - (s_m^\prime - s) \right\vert = \vert s_n^\prime - s_m^\prime \vert < \varepsilon,$$ showing that the sequence $\{ \vert s_n^\prime - s \vert \}$ is Cauchy. Therefore $\lim_{n \to \infty} \vert s_n^\prime - s \vert$ exists and we also have $$0 \leq \lim_{n \to \infty} \vert s_n^\prime - s \vert \leq \varepsilon$$ for every $\varepsilon >0$, which shows that the last limit is $0$ and therefore $$\lim_{n \to \infty} s_n^\prime = s.$$

Is this presentation correct? Have I understood Rudin's logic correctly?

If there are any errors or issues in my presentation, please do point those out!!

In proving this result, we have only used the axioms of a Banach space; so this result holds in any Banach space Am I right?

Does this result hold in every normed space?

• It looks like a bit too much. The sequence $(s_n'-s_n)$ converges to $0$, as per Rudin's argument. Since $(s_n)$ converges to $s$, we get immediately that $(s_n')$ converges to $s$. – David Mitra May 8 '16 at 17:35
• @DavidMitra yes, you're right. But I wanted to just use what Rudin has already demonstrated. I just wanted to stay confined within Rudin's framework. – Saaqib Mahmood May 8 '16 at 17:51