# A detailed proof of Riemann rearrangement theorem

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 $$(\alpha + M)/2 or $$(\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.

• 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...) – Sangchul Lee Sep 3 '19 at 0:02
• 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? – Akira Sep 3 '19 at 6:23
• 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. – Sangchul Lee Sep 3 '19 at 9:28
• Thank you so much @SangchulLee :) – Akira Sep 3 '19 at 9:32