# Question about series rearrangement in Baby Rudin (theorem 3.54).

I have trouble following the following theorem in Rudin:

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 $$\lim_{n\to\infty}\inf s_n^\prime = \alpha, \ \ \ \mbox{ and } \ \ \ \lim_{n\to\infty}\sup s_n^\prime = \beta. \tag{24}$$

Here's the 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 $$\sum p_n$$ 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).

I don't understand the last two lines of the proof (put in bold). I'm aware that this question was already asked on this forum, but I didn't understand the answers provided in this question, so that's why I'm writing my own question.

• I'm not sure what you want to number or I'd edit the question myself. You can number an equation as (24) by adding \tag{24} at the end of it. – saulspatz Jul 24 '18 at 15:29
• Thanks for your comment. I edited the post. – user370967 Jul 24 '18 at 15:34

First, let me make sure it's clear what's happening behind all those formulas. The rearrangement that allegedly works goes like this: First take just enough positive terms from your given series to produce a partial sum $>\beta$. (You can do that because the series of all the positive terms diverges.) After that, put just enough negative terms to bring the partial sum down below $\alpha$ (possible because the series of all the negative terms diverges). Then resume putting just enough positive terms to bring the partial sum back up above $\beta$. Continue working back and forth like this.

Notice that I said just enough terms at every stage. That ensures that, when you get a partial sum $s$ above $\beta$, it won't be too far above $\beta$; the difference $s-\beta$ will be at most the last term you added, because otherwise you could have stopped adding positive terms sooner. Similarly, when the partial sum goes below $\alpha$, the difference will be (in absolute value) at most the (absolute value of the) last term you added.

But your original series converged (conditionally), so the terms approach zero. That means that the amounts by which you overshoot $\beta$ and undershoot $\alpha$ are eventually arbitrarily small as you perform more and more stages of the process. And that's what those last two lines in Rudin's proof are saying is "clear".

I will consider the case $$\beta \in \mathbb{R}$$ here.
Rudin showed that $$x_n \to \beta$$ on p.77.
Let $$\epsilon$$ be an arbitrary positive real number.
Then, there exists a natural number $$N$$ such that $$n \geq N \implies \beta - \epsilon < x_n < \beta + \epsilon.$$

By the construction of $$\{s'_n\}$$, the follwoing inequalities hold:

$$\beta + \epsilon > x_N = s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N} > s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N + 1} \\> \cdots > s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N + k_N} = y_N < s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N + k_N+1} < \cdots < s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N + k_N+(m_{N+1}-1)} < s'_{m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N + k_N+m_{N+1}} = x_{N+1} < \beta + \epsilon.$$

Let $$M := m_1 + k_1 + \cdots + m_{N-1} + k_{N-1} + m_N$$.
The above inequalities say that $$s'_n < \beta + \epsilon$$ for all $$n \in \{M, M+1, \cdots, M + k_N + m_{N+1}\}$$.

And we see that the following ineqality holds:

$$s'_n < \beta + \epsilon$$ for all $$n \in \{M, M+1, \cdots\}$$.

• Thank you for your answer. I will look at it once I find time :) – user370967 Jan 30 '19 at 17:52