I'm trying to formulate the ideas from here and here into a rigorous argument. Honestly, this attempt takes me the entire three days to complete, and it is full of indexes of subscripts and inequalities. I feel that there is a very high chance that I made mistakes somewhere.

Could you please verify if my attempt is fine or it contains logical gaps/mistakes? I'm grateful for your help!

Theorem: Let $\sum x_{k}$ be a conditionally convergent series in $\mathbb{R}$ and $-\infty \leq \alpha \leq \beta \leq +\infty$. Then there exists a rearrangement $\sum x_{\sigma(k)}$ with partial sums $s_n = \sum_{k=0}^n x_{\sigma(k)}$ such that $$\liminf_{n \to \infty} s_n = \alpha, \quad \limsup_{n \to \infty} s_n = \beta$$

My attempt:

We define $x^+,x^-:\mathbb N \to \mathbb R$ and $\Delta^+,\Delta^-:\mathbb N \to \mathbb R$ by $$x^+_k = \frac{|x_k|+x_k}{2}, \quad x^-_k = \frac{x_k - |x_k|}{2}, \quad \Delta^+_n = \sum_{k=0}^n x^+_k, \quad \Delta^-_n = \sum_{k=0}^n x^-_k$$

Lemma: If $\sum x_k$ converges conditionally, then $\Delta^+$ is increasing and not bounded from above, and $\Delta^-$ is decreasing and not bounded from below.

We take sequences $(\alpha_n)_{n \in \mathbb N},(\beta_n)_{n \in \mathbb N}$ in $\mathbb R$ such that $\alpha_n \to \alpha,\beta_n \to \beta$ as $n \to \infty$ and $\alpha_m < \beta_n$ for all $m,n \in \mathbb N$. We define $\mu, \nu : \mathbb{N} \rightarrow \mathbb{N}$ recursively by $$\mu_0 = \min \{n \in \mathbb{N} \mid \Delta^+_n > \beta_0\}, \quad \nu_0 = \min \{n \in \mathbb{N} \mid \Delta^+_{\mu_0} + \Delta^-_n < \alpha_0\}$$ $$\mu_{p+1} = \min \{n \in \mathbb{N} \mid \Delta^+_n + \Delta^-_{\nu_p} > \beta_{p+1}\}, \quad \nu_{p} = \min \{n \in \mathbb{N} \mid \Delta^+_{\mu_{p}} + \Delta^-_n < \alpha_p\}$$

Because of the Lemma, $\mu$ and $\nu$ are well defined. By construction, we have $\Delta^+_{\mu_{p+1}} + \Delta^-_{\nu_p} > \beta_{p+1}$ and $\Delta^+_{\mu_{p}} + \Delta^-_{\nu_{p}} < \alpha_p$, so $\Delta^+_{\mu_{p+1}} > \beta_{p+1} - \Delta^-_{\nu_p} > \alpha_p - \Delta^-_{\nu_p} > \Delta^+_{\mu_{p}}$. As such, $\mu$ is strictly increasing. We also have $\Delta^+_{\mu_{p+1}} + \Delta^-_{\nu_p} > \beta_{p+1}$ and $\Delta^+_{\mu_{p+1}} + \Delta^-_{\nu_{p+1}} < \alpha_{p+1}$, so $\Delta^-_{\nu_{p+1}} < \alpha_{p+1} - \Delta^+_{\mu_{p+1}} <$ $\beta_{p+1} - \Delta^+_{\mu_{p+1}}$ $< \Delta^-_{\nu_p}$. As such, $\nu$ is strictly increasing.

By construction, $(\Delta^+_{\mu_{p+1}} +\Delta^-_{\mu_{p}}) - x^+_{\mu_{p+1}} \le \beta_{p+1}$ and $\alpha_p \le (\Delta^+_{\mu_{p}} + \Delta^-_{\nu_{p}}) - x^-_{\nu_{p}}$ for all $p \in \mathbb N$. Define $y : \mathbb{N} \rightarrow \mathbb{R}$ by taking $y_n$ to be the $n$th term in $$x^+_0, \ldots, x^+_{\mu_0}, x^-_0, \ldots, x^-_{\nu_0}, x^+_{\mu_0+1}, \ldots, x^+_{\mu_1}, x^-_{\nu_0+1}, \ldots, x^-_{\nu_1}, \ldots$$

Because $\mu$ and $\nu$ are strictly increasing, $y$ is a rearrangement of $x$. We define $s: \mathbb N \to \mathbb R$ by $$s_n = \sum_{k=0}^n y_k$$

It follows that $s_{\mu_p+ \nu_p +1} = \Delta^+_{\mu_{p}} + \Delta^-_{\nu_{p}}$ and $s_{\mu_{p+1} + \nu_p +1} = \Delta^+_{\mu_{p+1}} + \Delta^-_{\nu_{p}}$ for all $p \in \mathbb N$. Hence $\alpha_p - s_{\mu_p+\nu_p +1} \le - x^-_{\nu_{p}}$ and $s_{\mu_{p+1} + \nu_p +1} - \beta_{p+1} \le x^+_{\mu_{p+1}}$ for all $p \in \mathbb N$. As such, $|\alpha_p - s_{\mu_p + \nu_p +1}| \le - x^-_{\nu_{p}}$ and $|s_{\mu_{p+1} + \nu_p +1} - \beta_{p+1}| \le x^+_{\mu_{p+1}}$ for all $p \in \mathbb N$. On the other hand, $- x^-_{\nu_{p}} \to 0$ and $x^+_{\mu_{p+1}} \to 0$ as $p \to \infty$, so we have $$\lim_{p \to \infty} s_{\mu_p + \nu_p +1} = \lim_{p \to \infty} \alpha_p = \alpha, \quad \lim_{p \to \infty} s_{\mu_{p+1} + \nu_p +1} = \lim_{p \to \infty} \beta_{p+1} = \lim_{p \to \infty} \beta_{p} = \beta$$

As such, $$\liminf_{n \to \infty} s_n \le \lim_{p \to \infty} s_{\mu_p + \nu_p +1} = \alpha, \quad \beta = \lim_{p \to \infty} s_{\mu_{p+1} + \nu_p +1} \le \limsup_{n \to \infty} s_n$$

Assume that the sub-sequence $(s_{n_k})_{k \in \mathbb N}$ converges to $M < \alpha$. We have:

  1. There exists $\overline k \in \mathbb N$ such that $|s_{n_k} - M| < (\alpha-M)/2$ for all $k \ge \overline k$, so $s_{n_k} < (\alpha + M)/2$ for all $k \ge \overline k$.

  2. Because $s_{\mu_p + \nu_p +1} \to \alpha$ as $n \to \infty$, there exists $\overline p \in \mathbb N$ such that $|s_{\mu_p + \nu_p +1} - \alpha| < (\alpha-M)/2$ for all $p \ge \overline p$. Thus $(\alpha + M)/2 < s_{\mu_p + \nu_p +1}$ for all $p \ge \overline p$.

  3. If $k \in \mathbb N$ such that $n_k \ge \mu_1 + \nu_1 +1$, then there exists $f(k) \in \mathbb N$ such that $\mu_{f(k)} + \nu_{f(k)} +1 \le n_k \le \mu_{f(k)+1} + \nu_{f(k)+1} +1$. Moreover, $s_{n_k} \ge \min \{s_{\mu_{f(k)} + \nu_{f(k)} +1} ,s_{\mu_{f(k)+1} + \nu_{f(k)+1} +1}\}$.

  4. The set $A = \{k \in \mathbb N \mid k \ge \overline k \, \text{and} \, f(k)\ge \overline p\}$ is infinite. Let $K = \min A$.

It follows from $(1),(2),(3),(4)$ that at least one of the following inequalities holds $$(\alpha + M)/2 <s_{\mu_{f(K)} + \nu_{f(K)} +1} \le s_{n_K} < (\alpha + M)/2$$ or $$(\alpha + M)/2 <s_{\mu_{f(K)+1} + \nu_{f(K)+1} +1} \le s_{n_K} < (\alpha + M)/2$$

Both cases lead to a contradiction. As such, there is no sub-sequence of $(s_n)_{n \in \mathbb N}$ that converges to $M < \alpha$. Hence $$\liminf_{n \to \infty} s_n = \alpha$$

Similarly, we have $$\limsup_{n \to \infty} s_n = \beta$$

This completes the proof.

  • $\begingroup$ I am only at the halfway through reading the proof, and up to this point the only issue is the typo you made at defining $\Delta_n^-$ (you should sum $x_k^-$'s). Also, is there any particular reason to use $\alpha_n$'s and $\beta_n$'s where the constant sequences $\alpha_n=\alpha$ and $\beta_n=\beta$ should work equally? (I mean, it seems to me that allowing $\alpha_n$'s and $\beta_n$'s to vary does not add elegance to the proof...) $\endgroup$ – Sangchul Lee Sep 3 '19 at 0:02
  • $\begingroup$ Dear @SangchulLee, I allow $\alpha_n$'s and $\beta_n$'s vary to include the case $\alpha= -\infty$ and $\beta= +\infty$. Have you found any mistake in the remaining proof? $\endgroup$ – Akira Sep 3 '19 at 6:23
  • $\begingroup$ Ah, that makes sense now. Thank you :) As for the proofreading, I am afraid that I have no more comments as I have not read all the way to the end. Since the idea of the proof is so clear, however, I only suspect minor typos/gaps, if exist at all. $\endgroup$ – Sangchul Lee Sep 3 '19 at 9:28
  • $\begingroup$ Thank you so much @SangchulLee :) $\endgroup$ – Akira Sep 3 '19 at 9:32

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