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Let $(a_n)$ be a sequence of real numbers such that positive terms sum to $\infty$ and the negative terms sum to $-\infty$. Assume additionally that $\lim(a_n)=0$.

Is there a way to re-arrange the terms into a sequence $(b_n)$ such that $\sum b_n=0$?

My guess is yes, but I do not see why. I know that if we re-arrange $(a_n)$ we are guaranteed to get a sequence which also converges to $0$, but I do not see how to make sure the partial sums converge to $0$ as well.

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  • $\begingroup$ I don't have a proof, but my guess would be to rearrange as follows: [(largest positive term)+(smallest negative term)] + [(next largest >0 term)+(next smallest <0 term)] + .... This way the number in [ ] progressively gets smaller. $\endgroup$ – bjorn93 Oct 30 '19 at 23:01
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This is a conclusion of the Riemann rearrangement theorem.

In fact you can rearrange the sequence to get $\sum\limits_{n=0}^\infty b_n = M$ for an arbitrary $M \in \mathbb{R}$. The proof on wikipedia is quite nice to read.

The main idea is to partition $(a_n)$ into its positive and negative parts: $$ a_n^+ := \frac{a_n + |a_n|}{2},\ \ a_n^- := \frac{a_n - |a_n|}{2}$$

Now you take the first add enough $a_n^+$ so you overshoot 0:

$$\sum\limits_{n=1}^{p-1} a_n^+ \leq 0 < \sum\limits_{n=1}^p a_n^+$$ Now you add just enough $a_n^-$ to get below 0: $$\sum\limits_{n=1}^{p-1} a_n^+ +\sum\limits_{n=1}^{q} a_n^- < 0 \leq \sum\limits_{n=1}^p a_n^+ +\sum\limits_{n=1}^{q-1} a_n^-$$

You continue this by always slightly over/undershooting 0. Since $a_n \to 0$, the distance of your rearranged sum and 0 gets smaller and smaller until you finally converge.

Note: Why can you apply this theorem although it's normally stated for $\sum\limits_{n=0}^\infty a_n$ conditionally convergent? If you look into the proof it's important that $a_n \to 0$ (so that you get convergence) and that $\sum\limits_{k=0}^\infty a_n^+ = \infty$ and $\sum\limits_{k=0}^\infty a_n^- = -\infty$ (so that you can always overshoot/undershoot your value).

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