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.

  • $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.