Here's the statement of Theorem 3.54 in *Principles of Mathematical Analysis* by Walter Rudin, 3rd edition:

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 $$\lim_{n\to\infty}\inf s_n^\prime = \alpha, \ \ \ \mbox{ and } \ \ \ \lim_{n\to\infty}\sup s_n^\prime = \beta.$$ [ Rudin has numbered this set of inequalities as (24). ]

Now I've got a couple of questions regarding Rudin's proof:

First, when he has taken real-valued sequences $\{\alpha_n\}$, $\{\beta_n\}$ such that $\alpha_n \to \alpha$, $\beta_n \to \beta$. But then he has also required that $\alpha_n < \beta_n$ and $\beta_1 > 0$. Now is either of these two inequalities necessary for the proof to proceed, especially $\beta_1 > 0$?

Second, in the very last sentence Rudin states: "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)." How is this statement true? I mean how to explicitly verify this?

For those who haven't got a copy of Rudin on hand, I'll edit this question to reproduce Rudin's proof in its entirety.

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. [ In fact both these series diverge to $+\infty$. Am I right? ]

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

Choose real-valued sequences $\{ \alpha_n \}$, $\{ \beta_n \}$ such that $\alpha_n \rightarrow \alpha$, $\beta_n \rightarrow \beta$, $\alpha_n < \beta_n$, $\beta_1 > 0$. [ What if $\beta_1 \leq 0$? What if $\alpha_n \geq \beta_n$ for some $n$? What will go wrong? ]

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). [ How do we show this? ]