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This theorem has already been asked about, however my specific question, to my knowledge has not. I hope it will be acceptable if I refer the reader of this question to either here or here for Rudin's proof, since it has already been written up.

Rudin says that for "sequences $(m_n)$, $(k_n)$" the following is "clearly a rearrangement of $\Sigma a_n$.'' $$ P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} - Q_{k_1+1} - \cdots - Q_{k_2} + \cdots $$

Can anyone tell me, is he saying that $\Sigma a_{n'}$ is a rearrangement of $\Sigma a_n$ where $(a_{n'})$ is defined $$ a_{1'} := \sum_{i=1}^{m_1} P_i \hspace{1mm}-\hspace{1mm} \sum_{j=1}^{k_1} Q_j, \quad\ldots\quad a_{n'} := \sum_{i=m_{n'}+1}^{m_{{n'}+1}} P_i \hspace{1mm}-\hspace{1mm} \sum_{i=k_{n'}+1}^{k_{{n'}+1}} Q_j, \quad\ldots $$ for any sequences of integers $(m_n)$ and $(k_n)$? Or is he saying that $\Sigma a_{n'}$ is a rearrangement of $\Sigma a_n$ where $(a_{n'})$ is defined $$ a_1 := P_1, \quad\ldots\quad a_{n'} = \begin{cases} P_{\ell+1} &\text{ if }\hspace{2mm} \ell\ni a_{n'-1} = P_\ell \wedge \ell\neq m_n\hspace{1mm}\forall n\in\{1,2,\ldots\} \\ Q_{\ell+1} &\text{ if }\hspace{2mm} \ell\ni a_{n'-1} = Q_\ell \wedge \ell\neq k_n\hspace{1mm}\forall n\in\{1,2,\ldots\} \\ Q_{k_\ell} &\text{ if }\hspace{2mm} \exists\ell\ni a_{n'-1} = P_{m_\ell} \\ P_{m_\ell + 1} &\text{ if }\hspace{2mm} \exists\ell\ni a_{n'-1} = Q_{m_\ell} \end{cases} $$

Moreover, this only makes sense if $(m_n)$ and $(k_n)$ are strictly increasing sequences of positive integers (although Rudin does not make this specification), is that correct?

Thanks to anyone who is willing to help.

Edit: My suspicion is that the latter of these two is correct, but I would be grateful for a second or third opinion.

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    $\begingroup$ Doesn't the construction of $(m_n)$ and $(k_n)$ make them strictly increasing? $\endgroup$ – saulspatz Jul 9 at 23:16
  • $\begingroup$ I suppose so, yeah, that's a good point. Although, Rudin says that this is "clearly a rearrangement" before constructing $(m_n)$ and $(k_n)$. I think you're right though. $\endgroup$ – Thomas Winckelman Jul 9 at 23:46
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    $\begingroup$ I don't think this proof was one of Rudin's better efforts. In my view, it obscures the essential idea under a welter of notation. $\endgroup$ – saulspatz Jul 9 at 23:51
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    $\begingroup$ @saulspatz Rudin wants to prove it all in one shot. Better to prove one thing (really) clearly, say rearrangements can converge to any number, and then state his general theorem without proof... $\endgroup$ – CopyPasteIt Jul 10 at 13:05
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Rudin is building a rearrangement of the $a_n$ terms via the recursion theorem:

Theorem 1: Let $X$ be a set with $x_0 \in X$ and $\psi: X \to X$ be any function. Then there exist one and only one function $\rho: \Bbb N \to X$ satisfyings

$\tag 1 \rho(0) = x_0$ $\tag 2 \forall n \in \Bbb N, \; \rho(n+1) = \psi(\rho(n))$

See, for example, the wikipedia article recursive definition.

You won't find recursion in the index. Apparently, for Rudin, building sequences via recursion is so routine it doesn't deserve mention. If you want to learn it, read his proofs!

You will notice that in the body of the proof you won't find any attempt to 'map things out' - the recursion machine implicitly takes care of things at 'run time'.

I suggest the serious student rework the proof for three cases:

Case 1: For all $n$, $\alpha_n = \alpha \in \Bbb R$ and $\beta_n = \beta \in \Bbb R$ and $\alpha = \beta$.

Case 2: For all $n$, $\alpha_n = \alpha \in \Bbb R$ and $\beta_n = \beta \in \Bbb R$ and $\alpha \lt \beta$.

Case 3: For all $n$, $\alpha_n = -n$ and $\beta_n = +n$.

Try working on Case 1 by first thinking it through and working out your ideas on some scrap paper.

Notice that in Rudin's proof he wants $\beta_1 \gt 1$. No doubt he needs that to build (seed) his machine. It is a red herring, but if you understand that you will be in business.

Finally, to really understand the theorem you can use Rudin's proof to get the main ideas and then build your own recursion machine(s).

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    $\begingroup$ I like that a lot. I'm going to try and take your idea and run with it, thank you so much. I hope to put together a proof of the kind you're describing within the next few days. By the way you're 100% right about Rudin not saying what he is doing. To be honest, I always figured that he was outlining a proof by induction of the existence of a sequence when, in actuality, he was defining a sequence recursively. $\endgroup$ – Thomas Winckelman Jul 10 at 22:31
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    $\begingroup$ @ThomasWinckelman The definition of a rearrangement makes it appear that you have to set out and define $k_n$ so that ${a_n}' = a_{k_n}$, but really you just want to build ${a_n}'$, with the corresponding $k_n$ being implicitly built at the same time; you can throw $k_n$ away. $\endgroup$ – CopyPasteIt Jul 10 at 22:43
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    $\begingroup$ @ThomasWinckelman Yes, I've mixed up induction with recursive definition at times. $\endgroup$ – CopyPasteIt Jul 10 at 22:48
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    $\begingroup$ @ThomasWinckelman Also, listen to to Rudin in Example 3.53 and finish it up. Even though here you can explicitly/upfront describe $k_n$ to get $1 + 1/3 - 1/2 + 1/5 + \dots$, you can also define it with recursion as in Theorem 3.54. $\endgroup$ – CopyPasteIt Jul 10 at 22:55
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Looks to me like it's close to the latter option. (As saulspatz commented, $(m_n)$ and $(k_n)$ are constructed to be strictly increasing, so that's not an issue.) The point here is that each of the terms $P_i$ is by definition one of the terms $a_n$, and each of the terms $-Q_j$ is by definition one of the terms of $a_n$, and they each correspond to a distinct term, so a sum of all the $P_i$s and $-Q_j$s is a rearrangement of $\sum a_n$.

The precise definition of the sequence $a'_n$ would be as follows, adopting the convention $m_0=k_0=0$:

$$ a'_n = \begin{cases}P_{n-k_j} &\text{if } m_j+k_j < n \leq m_{j+1}+k_j \\ -Q_{n-m_j} &\text{if } m_j+k_{j-1}<n\leq m_j+k_j \end{cases} $$

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  • $\begingroup$ Thank you I appreciate it! $\endgroup$ – Thomas Winckelman Jul 10 at 22:40
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for what it's worth, I finally put together a pretty thorough reworking of Rudin's proof.

I'd be happy to receive feedback, but mostly I just thought I'd share in case anyone cares. Here's a link to the PDF, because I don't know how to make my LaTeX macros compatible with this website (sorry).

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