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?

  • $\begingroup$ 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 $s$. $\endgroup$ – David Mitra May 8 '16 at 17:35
  • $\begingroup$ @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. $\endgroup$ – Saaqib Mahmood May 8 '16 at 17:51

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