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.
 A: 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).
