# Question about Theorem 3.54 Rudin.

Let $$\sum a_n$$ be a conditionally convergent series. Suppose : $$-\infty \leq \alpha\leq \beta \leq +\infty$$ Then there exist a rearrangement $$\sum a'_n$$ with partial sums $$\{s'_n\}$$ such that

$$\lim\limits_{n\to\infty} \inf s'_n=\alpha \quad \quad \lim\limits_{n\to\infty}\sup s'_n=\beta$$ 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 $$\sum p_n$$ and 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).

Why is $$| x_n - \beta_n | \leq P_{m_n}$$? Why $$P_n\to 0$$? What if $$\sum a_n$$ converges absolutely?

I don't know how this inequality is true but makes sense because it says something like "By definition $$m_n$$ is the smallest integer such that $$x_n > \beta_n$$" which might be key to it. I think $$P_n\to 0$$ is because $$a_n\to 0$$ but it is still not clear since there might be exception. $$\sum a_n$$ shouldn't converge absolutely for this theorem makes sense but I don't know why.

By definition, $$x_n>\beta_n$$. If $$x_n-\beta_n>P_{m_n}$$, then $$x_n-P_{m_n}>\beta_n$$, contradicting the choice of $$m_n$$.
The sequence $$\{P_n\}$$ is a subsequence of $$\{a_n\}$$; as the latter goes to zero, so does the former.
If the series converges absolutely then any rearrengement converges to the same limit. So in that case the result can only be possible when $$\alpha=\beta=\sum_na_n$$.
To see what happens with $$\sum_n|a_n|<\infty$$, let $$\sigma:\mathbb N\to\mathbb N$$ be a bijection. Let $$L=\sum_na_n$$. Fix $$\varepsilon>0$$. By hypothesis, there exists $$n_0$$ such that $$\sum_{n>n_0}|a_n|<\varepsilon$$. Let $$F=\sigma^{-1}(\{1,\ldots,n_0\})$$ and put $$m=\max\{k:\ k\in F\}$$. Then $$\Bigl|L-\sum_{k=1}^ma_{\sigma(k)}\Bigr|=\Bigl|\sum_{k>m}a_{\sigma(k)}\Bigr| \leq \sum_{k>m}|a_{\sigma(k)}|\leq\sum_{k>n_0}|a_k|<\varepsilon,$$ where the second to last inequality is justified by the fact that if $$k>m$$, then $$k\not\in F$$ and so $$\sigma(k)\not\in\{1,\ldots,n_0\}$$, that is $$\sigma(k)>n_0$$.
• I do not like Rudin's presentation. Let $f(1)=a_1.$ Let $S_n=\sum_{j=1}^na_{f(j)}.$ Let $U(n+1)=\min \{m\not \in \{f(j):j\le n\}: a_m\ge 0\}$ and let $V(n+1)=\min \{m\not \in \{f(j):j\le n\}: a_m< 0\}.$ Let $f(n+1)=U(n+1)$ if $S_n\le \alpha$ or if $[\alpha<S_n< \beta \land a_{f(n)}\ge 0].$ Otherwise let $f(n+1)=V(n+1).$ Then $f:\Bbb N\to\Bbb N$ is bijective & $\lim\inf S_n=\alpha$ & $\lim\sup S_n=\beta.$ Commented Jan 6, 2021 at 21:49
• ERRATUM.In the 1st line of my comment above, that should say: Let $f(1)=1$. Commented Jan 6, 2021 at 21:58