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In the infinite series chapter of Spivak's Calculus, there is a theorem regarding rearrangements:

If $\sum_{n=1}^{\infty} a_n$ converges, but does not converge absolutely, then for any number $\alpha$, there is a rearrangement $\{b_n\}$ of $\{a_n\}$ such that $\sum_{n=1}^{\infty}b_n = \alpha.$

The proof presented in the book is as follows: Let $\alpha > 0$. Let $\sum_{n=1}^{\infty}p_n$ denote the series formed by the positive terms of $\{a_n\}$, and let $\sum_{n=1}^{\infty}q_n$ denote the series formed by the negative terms. Because $\sum_{n=1}^{\infty} a_n$ is not absolutely convergent, both $\sum_{n=1}^{\infty}p_n$ and $\sum_{n=1}^{\infty}q_n$ do not converge, and hence are unbounded.

Therefore, let $N_1$ be the least integer such that $\sum_{n=1}^{N_1} p_n > \alpha.$ Set $S_1 = \sum_{n=1}^{N_1} p_n.$ Because $N_1$ is the least integer, $S_1 - p_{N_1} \leq \alpha,$ so $$S_1- \alpha \leq p_{N_1}.$$

Now, let $M_1$ be the least integer such that $S_1 + \sum_{n=1}^{M_1} q_n < \alpha.$ Set $T_1 = S_1 + \sum_{n=1}^{M_1} q_n.$ Because $M_1$ is the least integer, $T_1 - q_{M_1} \geq \alpha,$ so $$\alpha - T_1 \leq -q_{M_1}.$$

Continue this procedure indefinitely, alternating between sums larger and smaller than $\alpha$, each time using appropriately small $N_k$ and $M_k$. From this, we obtain $$\{ b_n \} = p_1, p_2, ... , p_{N_1}, q_1, ...., q_{M_1}, p_{N_1 + 1},... , p_{N_2}, ...$$

By construction, $|S_k - \alpha| \leq p_{N_k},$ and $|\alpha - T_k| \leq -q_{M_k}.$ Since $\lim_{n \to \infty} a_n = 0$, it follows that the series of our rearrangement $\{ b_n \}$ of $\{ a_n \}$ converges to $\alpha$.

I have two questions about this proof:

  1. The last line of the proof (bolded) seems pretty handwavy, but I've been unable to find a rigorous argument for it. It seemed like some kind of subsequence argument on the sequence of partial sums of $\{ b_n \}$...
  2. What is the motivation behind this rearrangement construction?

Thank you so much for your help!

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    $\begingroup$ $p_{N_k}$ and that $q_{M_k}$ are terms of the sequence $(a_n)$ and for a convergent series $\sum a_n$, $a_n\to 0$ so for sufficiently large k, $p_{N_k}$ also tends to 0. $\endgroup$
    – Koro
    May 29 '20 at 7:32
  • $\begingroup$ You may find this answer useful. $\endgroup$
    – Paramanand Singh
    May 29 '20 at 15:07
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As mentioned in the comments, since the series $\sum_n a_n$ converges, the individual summands must go to $0$, $a_n \to 0$. If I recall correctly, Spivak refers to this as the "vanishing criterion" or something like that.

Now, the $p_n$'s and $q_n$'s are a subsequence of the $a_n$. Recall that if a sequence $a_n \to 0$, then every subsequence also converges to $0$; in particular both $p_n \to 0$ and $q_n \to 0$ (and $p_{N_k} \to 0$ and $-q_{M_k} \to 0$ as $k \to \infty$, since subsequence of a subsequence is still a subsequence, and hence will also converge to $0$). Now, the bolded sentence can be viewed as an application of the squeeze theorem for sequences. Or just show it directly:

Given $\epsilon > 0$, choose $\nu\in \Bbb{N}$ large enough so that $k> \nu$ implies $0 \leq |p_{M_k}| < \epsilon$. Such a $\nu$ exists since $\lim_{k \to \infty} p_{M_k} = 0$. Therefore, we have $|S_k - \alpha| \leq p_{M_k} < \epsilon$. Since $\epsilon > 0$ is arbitrary, it completes the proof that $\lim_{k \to \infty} S_k = \alpha$.

A similar proof holds for the $q$'s.


Finally, the motivation for the proof is pretty straight forward. You know that both the positive and negative series do not converge. So, $\sum_n p_n = \infty$ and $\sum_n q_n = - \infty$. So, the idea is that given any $\alpha$, to construct the rearrangement, you take a bunch of positive terms $p$ so that the sum just barely exceeds $\alpha$, then you take a bunch of negative terms so that the total sum is just below $\alpha$, then take a bunch of positive terms, then a bunch of negative ones. Each time, the sequence of partial sums will be hopping to the right and left of $\alpha$, and the "amount by which it hops" will eventually decrease, so that the limit of the partial sums of the rearrangement actually converges to $\alpha$.

Imagine letting a pendulum swing. First, it will overshoot and swing to the right, then it will swing to the left, then it will swing to the right, and then left again, but eventually, the amplitudes of oscillation will get smaller and smaller, until it eventually stabilizes. That's pretty much the illustration to keep in mind for this proof. (In this analogy, the motion of the pendulum is like the partial sums of the rearrangement)

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  • $\begingroup$ That pendulum analogy is so cool! Thank you! $\endgroup$
    – alexion
    May 29 '20 at 10:03
  • $\begingroup$ @alexion haha you're welcome. in fact when i first studied this theorem, I didn't really understand the proof, but now after answering this question I think i can never forget the theorem or the proof lol. So thanks for the question as well $\endgroup$
    – peek-a-boo
    May 29 '20 at 17:10
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    $\begingroup$ @alexion by the way a word of warning: the analogy with the pendulum isn't quite perfect, because a real pendulum (due to friction) oscillates with strictly decreasing amplitude. i.e you can never go "higher/further" than you previously went. However, for this proof, notice that it is nowhere stated (and in fact in general false) that $p_{N_1} \geq p_{N_2} \geq p_{N_3} \geq \dots \geq 0$. i.e the sequence of $p$'s is not monotonically decreasing; all we know is that the limit is zero. So, sometimes, it's possible for the partial sums to exceed what it was previously. $\endgroup$
    – peek-a-boo
    May 29 '20 at 21:26
  • $\begingroup$ but that doesn't matter, because all we care about is that the partial sums converge to $\alpha$, and as the proof shows quantitatively, this is indeed the case. So basically my point is that you should take the pendulum analogy with a grain of salt, and keep in mind this small caveat. $\endgroup$
    – peek-a-boo
    May 29 '20 at 21:28
  • $\begingroup$ yes, the width of the oscillations (partial sums) decrease over time, but no guarantee that the next oscillation width will be less than the one before, right? $\endgroup$
    – alexion
    May 30 '20 at 1:34

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