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? ]

  • 5
    $\begingroup$ The way I see, the inequalities $\alpha_n<\beta_n$ and $\beta_1>0$ are not really necessary, but they make the proof easier to visualize. For the second question, let $\epsilon>0$. Then $\exists n_0$ s.t. for all $n\geq n_0$, $\beta_n$ is $\epsilon$-close to $\beta$, and $P_n<\epsilon$. The rearrangement $(a'_m)$ is defined in such a way that sufficiently large partial sums, are at most $\beta_n+P_n$, which is less than $\beta+2\epsilon$ for $n\geq n_0$. Similarly, sufficiently large partial sums are at least $\alpha_n-Q_n$, which is close to $\alpha$ for $n$ large. $\endgroup$ Nov 26, 2014 at 15:13

3 Answers 3


This is a theorem with a direct proof by construction.

The statement simply says "there exists a rearrangement". So the book assembles one. And it does not mean there cannot exist any other construction! The proof states one case, namely expression (25) and shows it works. That's it.

The structure of the proof is basically the following.

First, we put up Expression (25). Then we show that all its constituents indeed exist and make sense, namely $p_n, q_n, P_{m_k}, Q_{m_k}, \beta_n, \alpha_n$.

A picture is worth a thousand words.

enter image description here

Curves in blue and in green are sequences $\beta_n$ and $\alpha_n$, they converge to $\beta$ and $\alpha$ respectively.

Curve in red is our Expression (25). The way we painted this picture shows a few things about Expression (25).

The first terms of it are positive, so the reason why $\beta_1 >0$, and the curve first goes up. Again, in our construction we first put Expression (25) as a basis, starting point of the proof. The first terms of it are positive, so $\beta_1$ must be positive. We tweak betas and other components to make them tie in with Expression (25).

Next, $x_n > \beta_n$ and $y_n < \alpha_n$, so the red curve of Expression (25) sticks out a bit "outside" sequences $\beta_n$ and $\alpha_n$. By one last positive term and negative term $P_{m_n}$ and $Q_{k_n}$.

Differences between Expression (25) and these sequences (marked X) become smaller because $P_n \to 0$ and $Q_n \to 0$. It is precisely for this reason it becomes finally "clear" that Expression (25) cannot converge to any number greater than $\beta$ or smaller than $\alpha$.

Can we have an alternative construction? Why not!

enter image description here

Just change Expression (25), and tweak all other elements accordingly, e.g. make Expression (25) "expand" inside the "space" between $\beta$ and $\alpha$. Make the first terms negative numbers, the curve would first go down in this case. The differences (marked Y) would also gradually become smaller. And the whole thing would work as well.

You might ask why the theorem is stated only for the real numbers? That's because all inequalities do not make sense for the complex numbers, which do not have '<' defined.

  • $\begingroup$ Thank you for your answer! We can allow $\beta_1=0$ too, right? Then the LHS of $P_1 + \cdots + P_{m_1} > \beta_1$ would simply be $P_1$, since $P_i>0$ by definition. Since $m_1=1$, we would also have that $|x_1-\beta_1|=P_1\le P_{m_1}$ holds. $\endgroup$
    – psie
    May 21, 2023 at 12:00

$\alpha_n<\beta_n$ and $\beta_1>0$

Those conditions are necessary to ensure the partial sums $x_n$ and $y_n$ are never equal and even that $x_n > y_n$.

$\beta_1>0$ guarantees the left part of the inequality $P_1 + \ldots + P_{m_1} > \beta_1$ contains at least one such $P_i > 0$ (See $x_1 = P_1 + \ldots + P_{m_1}$ as $y_0 + P_1 + \ldots + P_{m_1}$ with $y_0 = 0$).

While $\alpha_n<\beta_n$ ensures for $n \geq 1$ that, we must at least subtract a $Q_i$ from $x_n$ in order to obtain $y_n$.

Why have we to ensure $x_n > y_n$? Because if not, there no non-negative "smallest integer such that" each time (the sum may already satisfy the condition) and the construction is no more effective.


Here's a try: Let partial sums of $(25)$ be denoted by $S_v$ for $v\in\mathbb{N}$. Now if $m_i+k_i\le v<m_{i+1}+k_i$ then $y_i\le S_v<x_{i+1}$, and in other case when $m_i+k_{i-1}\le v<m_{i}+k_i$ then $y_i<S_v\le x_{i}$. Therefore we have $y_j\le S_v$. Now taking limit infimum on both sides we get $\liminf y_j\le\liminf S_v$, but we have shown that $y_n\rightarrow \alpha$ thus $\alpha\le\liminf S_v$. Similarly $\liminf S_v\le\beta$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .