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This question is about theorem 3.54 of Baby Rudin:

Statement and proof of that theorem are given here:

Theorem 3.54: Let $\sum a_n$ be a series of real numbers which converges, but not absolutely. Suppose $$-\infty \leq \alpha \leq \beta \leq +\infty.$$ Then there exists a rearrangement $\sum a_n^\prime$ with partial sums $s_n^\prime$ such that $$\lim_{n\to\infty}\inf s_n^\prime = \alpha, \ \ \ \mbox{ and } \ \ \ \lim_{n\to\infty}\sup s_n^\prime = \beta. \tag{24}$$

Proof: Let $$p_n = \frac{|a_n| + a_n}{2}, \ q_n = \frac{|a_n| - a_n}{2} \,\,\,\,\, (n = 1, 2, 3, \ldots). $$ Then $p_n - q_n = a_n$, $p_n + q_n = |a_n|$, $p_n \geq 0$, $q_n \geq 0$. The series $\sum p_n$, $\sum q_n$ must both diverge.

For if both were convergent, then $$\sum \left( p_n + q_n \right) = \sum |a_n|$$ would converge, contrary to hypothesis. Since $$ \sum_{n=1}^N a_n = \sum_{n=1}^N \left( p_n - q_n \right) = \sum_{n=1}^N p_n - \sum_{n=1}^N q_n,$$ divergence of $\sum p_n$ or convergence of $\sum q_n$ (or vice versa) implies divergence of $\sum a_n$, again contrary to hypothesis.

Now let $P_1, P_2, P_3, \ldots$ denote the non-negative terms of $\sum a_n$, in the order in which they occur, and let $Q_1, Q_2, Q_3, \ldots$ be the absolute values of the negative terms of $\sum a_n$, also in their original order.

The series $\sum P_n$, $\sum Q_n$ differ from $\sum p_n$, $\sum q_n$ only by zero terms, and are therefore divergent.

We shall construct sequences $\{m_n \}$, $\{k_n\}$, such that the series $$ P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} \\- Q_{k_1 + 1} - \cdots - Q_{k_2} + \cdots, \tag{25}$$ which clearly is a rearrangement of $\sum a_n$, satisfies (24).

Choose real-valued sequences $\{ \alpha_n \}$, $\{ \beta_n \}$ such that $\alpha_n \rightarrow \alpha$, $\beta_n \rightarrow \beta$, $\alpha_n < \beta_n$, $\beta_1 > 0$.

Let $m_1$, $k_1$ be the smallest integers such that $$P_1 + \cdots + P_{m_1} > \beta_1,$$ $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} < \alpha_1;$$ let $m_2$, $k_2$ be the smallest integers such that $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} > \beta_2,$$ $$P_1 + \cdots + P_{m_1} - Q_1 - \cdots - Q_{k_1} + P_{m_1 + 1} + \cdots + P_{m_2} - Q_{k_1 + 1} - \cdots - Q_{k_2} < \alpha_2;$$ and continue in this way. This is possible since $\sum P_n$, $\sum Q_n$ diverge.

If $x_n$, $y_n$ denote the partial sums of (25) whose last terms are $P_{m_n}$, $-Q_{k_n}$, then $$ | x_n - \beta_n | \leq P_{m_n}, \ \ \ |y_n - \alpha_n | \leq Q_{k_n}. $$ Since $P_n \rightarrow 0$, $Q_n \rightarrow 0$ as $n \rightarrow \infty$, we see that $x_n \rightarrow \beta$, $y_n \rightarrow \alpha$.

Finally, it is clear that no number less than $\alpha$ or greater than $\beta$ can be a subsequential limit of the partial sums of (25).

From chatroom, I got to know that $\beta_1>0$ (in fact, we can make it $\beta_1\ge 0$) is needed to ensure $|x_n-\beta_1|\le P_{m_1}$, and by $\alpha_n<\beta_n$, we ensure that $\alpha_n<x_n$ because by construction, $\beta_n<x_n$ ensuring that we need to subtract some $Q$'s to make $y_n<\alpha_n$.

My question is: What guarantees us that we need to add at least one $P$ to go from, e.g., $P_1+\ldots+P_{m_1}-Q_1-\ldots-Q_{k_1}<\alpha_1$ to $P_1+\ldots+P_{m_1}-Q_1-\ldots-Q_{k_1}+P_{m_1+1}+\ldots+P_{m_2}>\beta_2$? I mean $\beta_2$ can be so small that we don't need to add $P_{m_1+1}+\ldots+P_{m_2}$. And in this case, I think, expression $|x_2-\beta_2|\le P_{m_2}$ becomes meaningless.

Also, another question: Can't we accomplish our goal to have $\alpha_n<x_n$ by just $\alpha_n=\beta_n$ instead of $\alpha_n<\beta_n$?

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  • $\begingroup$ Thank you for your post. When you write $|x_n-\beta_1|\le P_{m_1}$, do you mean $|x_1-\beta_1|\le P_{m_1}$? $\endgroup$
    – psie
    May 20, 2023 at 12:25

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My question is: What guarantees us that we need to add at least one $P$ to go from, e.g., $P_1+\ldots+P_{m_1}-Q_1-\ldots-Q_{k_1}<\alpha_1$ to $P_1+\ldots+P_{m_1}-Q_1-\ldots-Q_{k_1}+P_{m_1+1}+\ldots+P_{m_2}>\beta_2$? I mean $\beta_2$ can be so small that we don't need to add $P_{m_1+1}+\ldots+P_{m_2}$. And in this case, I think, expression $|x_2-\beta_2|\le P_{m_2}$ becomes meaningless.

Your concern is valid. The proof requires adding the additional condition $a_{n-1} < b_{n}$ for the $\{a_n\}$ and $\{b_n\}$ sequences. This point has been noted in a list of errata for the text.

Also, another question: Can't we accomplish our goal to have $\alpha_n<x_n$ by just $\alpha_n=\beta_n$ instead of $\alpha_n<\beta_n$?

This wouldn't work because we need $\{x_n\} \to \beta$ and $\{y_n\} \to \alpha$. As you point out, the purpose of the condition $\alpha_n< \beta_n$ is to ensure that some $Q$'s get subtracted each iteration. We additionally need the $\alpha_{n-1} < \beta_{n}$ condition to ensure that some $P$'s always get added.

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    $\begingroup$ Thank you so much for taking pain in reading and replying. I still can't understand why $\alpha_n=\beta_n$ will not work: We get $x_n>\beta_n$ by construction, and we need $x_n>\alpha_n$. So, $\alpha_n=\beta_n$ will be as good as $\alpha_n<\beta_n$ , right? $\endgroup$
    – Silent
    Feb 18, 2018 at 18:23
  • $\begingroup$ But how would that work when $\alpha < \beta$? $\endgroup$
    – David
    Feb 19, 2018 at 1:23
  • $\begingroup$ Oh, alright! So, $\alpha_n<\beta_n$ covers all the cases, while with $\alpha_n=\beta_n$ we had to add unnecessary clauses. Thanks! $\endgroup$
    – Silent
    Feb 19, 2018 at 3:38

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