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Here's Theorem 3.55 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd 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.

Now 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 $$ \sum_{i =n}^m \left\vert a_i \right\vert \leq \varepsilon.$$ [This relation Rudin calls (26). ] Now choose $p$ such that the integers $1, 2, \ldots, N$ are all contained in the set $k_1, k_2, \ldots, k_p$ (we use the notation of Definition 3.52). Then if $n > p$, the numbers $a_1, \ldots, a_N$ will cancel in the difference $s_n - s_n^\prime$, so that $\left\vert s_n - s_n^\prime \right\vert \leq \varepsilon$, by (26). Hence $\left\{ s_n^\prime \right\}$ converges to the same sum as $\left\{ s_n \right\}$.

And, finally, here's Rudin's Definition 3.52.

Let $\left\{ k_n \right\}$, $n = 1, 2, 3, \ldots$, be a sequence in which every positive integer appears once and only once (that is, $\left\{ k_n \right\}$ is a 1-1 function from $J$ onto $J$, in the notation of Definition 2.2). Putting $$ a_n^\prime = a_n \ \ \ (n= 1, 2, 3, \ldots),$$ we say that $\sum a_n^\prime$ is a rearrangement of $\sum a_n$.

And, for the sake of completeness, Rudin uses the symbol $J$ to denote the set of natural numbers.

Now my question is, how does Rudin's relation (26) give the conclusion that $\left\vert s_n - s_n^\prime \right\vert \leq \varepsilon$ if $n > p$?

Here's how I have been able to understand the proof.

Given that $\sum \left\vert a_n \right\vert$ converges, we can conclude that $\sum a_n$ converges too. Let $$s = \sum_{n =1 }^\infty a_n.$$ Let $s_n$, $(n = 1, 2, 3, \ldots)$, be the partial sums of $\sum a_n$. Then $$s = \lim_{n \to \infty} s_n.$$ Now let $\sum a_n^\prime$ be a rearrangement of $\sum a_n$, and let $s_n^\prime$, $(n = 1, 2, 3, \ldots)$, be the partial sums of $\sum a_n^\prime$.

We show that $$\lim_{n \to \infty} s_n^\prime = s$$ as well. Now as $s_n \to s$ as $n \to \infty$, so, given $\varepsilon > 0$, we can find a natural number $N_1$ such that $$ \left\vert s_n - s \right\vert < \frac{\varepsilon}{2}$$ for all $n \in \mathbb{N}$ such that $n > N_1$.

Now as $\sum \left\vert a_n \right\vert$ converges, so we can find a natural number $N_2$ such that $$ \sum_{i =n}^m \left\vert a_i \right\vert < \frac{\varepsilon}{2}$$ for all $m, n \in \mathbb{N}$ such that $m \geq n \geq N_2$. So we can conclude that $$\left\vert \sum_{i=n}^m a_i \right\vert < \frac{\varepsilon}{2}$$ for all $m, n \in \mathbb{N}$ such that $m \geq n \geq N_2$.

Now let $N = \max \left\{ N_1, N_2 \right\}$. Then, for all $m, n \in \mathbb{N}$ such that $m \geq n > N$, we have $$ \left\vert s_n - s \right\vert < \frac{\varepsilon}{2}$$ and also $$\left\vert \sum_{i=n}^m a_i \right\vert < \frac{\varepsilon}{2}.$$

Now let $p$ be a natural number such that the integers $1, 2, \ldots, N$ are all contained in the set $\left\{ k_1, \ldots, k_p \right\}$. Then, for all $n \in \mathbb{N}$ such that $n > p$, we see that the difference $s_n - s_n^\prime$ is a sum of some finitely many terms of the sequence $\left( a_{N+1}, a_{N+2}, a_{N+3}, \ldots\right)$, and therefore $$\left\vert s_n - s_n^\prime \right\vert < \frac{\varepsilon}{2}. $$

So if $n \in \mathbb{N}$ is such that $n > \max \{ N, p \}$, then we have $$ \left\vert s_n - s_n^\prime \right\vert < \frac{\varepsilon}{2}$$ and also $$\left\vert s_n - s \right\vert < \frac{\varepsilon}{2}.$$ Therefore, for all $n \in \mathbb{N}$ such that $n > \max \{ N, p \}$, we have $$ \left\vert s_n^\prime - s \right\vert \leq \left\vert s_n^\prime - s_n \right\vert + \left\vert s_n - s \right\vert < \varepsilon,$$ from which it follows that $$ \lim_{n \to \infty} s_n^\prime = s$$ also.

Is my understanding of the proof of Theorem 3.55 in Baby Rudin correct? If so, then is my version the same as Rudin's? If not, where have I erred?

And, if my proof is also correct but differs from Rudin's, can anybody here please fill in the details in Rudin's original proof for me? Thanks.

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  • $\begingroup$ Looks valid, I didn't spot any errors. (I didn't carefully scrutinize it, so I may have overlooked some small glitch.) See here for how I think Rudin intended the proof. $\endgroup$ Oct 23, 2016 at 20:05

2 Answers 2

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Warm-up.

By Theorem 3.45, if a series converges absolutely, then the original series converges as well. Furthermore $\lim_{n\to\infty}a_n = 0$.

If a series converges, then for such series Cauchy criteria is true, that is Theorem 3.22. There exists $N$ such that if $m \ge n \ge N$ implies

$$ \left| \sum_{k=n}^{m} a_k \right| < \epsilon $$

In the above expression indexes $m, n$ are all finite despite the fact that any series contains infinite number of terms.

Abstract for Theorem 3.55

In order to prove our theorem we have to apply to the definition of a limit (a bit modified for two series). We have to show that by Definition 3.1 for any arbitrary $\epsilon >0$ there exists $N$ such that $n > N$ implies $d(a_n, b_n) < \epsilon$. In our case $a_n$ and $b_n$ are the series $s_n$ and $s'_n$.

Again, all we need to find is this $N$ such that $$ n \ge N \qquad \text{implies} \qquad |s_n - s'_n| < \epsilon $$

In our case it is $p$, index of the second series.

First, for the first series we have to find such $N$ so that

$$ m \ge k \ge N \qquad \text{implies} \qquad \left| \sum_{n=k}^{m} a_n \right| < \frac{\epsilon}{2} $$

Then, for this $N$ found for the first original series, in the rearranged series we find index $p$ such that the first $p$ terms of the second series include the first $N$ terms of the first series.

Why do we do this? We do this because we need the difference between these two series to be arbitrary small.

We have $$ \left|\underbrace{s'_n}_\text{$\le \frac{\epsilon}{2}$, sum of remaining terms} \quad - \quad \underbrace{s_n}_\text{$\le \frac{\epsilon}{2}$, sum of remaining terms} \right| < \epsilon $$

In the above expression the first N terms of $a_n$ cancel out, and both remaining bits are less than $\frac{\epsilon}{2}$.

To re-iterate, index $p$ of $s'_n$ is what we need for the arbitrary difference to be less than epsilon.

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$$N \leq p < n.$$ $$s_n = a_1 + \cdots + a_N + a_{N+1} + \cdots + a_n.$$ $$s'_n = a_{k_1} + \cdots + a_{k_N} + \cdots + a_{k_p} + a_{k_{p+1}} + \cdots + a_{k_n}.$$ $$\{1, \cdots, N\} \subset \{1, \cdots, n\}.$$ $$\{1, \cdots, N\} \subset \{k_1, \cdots, k_p\} \subset \{k_1, \cdots, k_n\}.$$ $$\{1,\cdots,N\}\subset\{k_1,\cdots,k_n\}\cap\{1,\cdots,n\}.$$ $$T:=(\{k_1,\cdots,k_n\}\cup\{1,\cdots,n\})-(\{k_1,\cdots,k_n\}\cap\{1,\cdots,n\}).$$ $$T \cap \{1, \cdots, N\} = \emptyset.$$ $$\text{If } i \in T, \text{ then } i \geq N+1.$$ $$|s_n - s'_n| \leq \sum_{i \in T} |a_i| \leq \sum_{i=N+1}^{\max{T}} |a_i| \leq \epsilon.$$

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