# Let $a_i$ satisfy $\lim(a_n)=0$ and have subsequences whose partial sums diverge to $\infty$ subsequence $(b_n)$ such that $\sum_{n=1} ^ \infty b_n=0$

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.

• 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. – 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).