Regd. Riemann Rearrangement Assume that A is an arbitrary set and there exists a bijection $\phi : B \rightarrow A$ and $x_{\alpha} \in [0,\infty]$, the book says that 
$$\Sigma_{\alpha \in A} x_{\alpha}= \Sigma_{\beta\in B}  x_{\phi(\beta)}$$
can fail if the series is not absolutely convergent and we are dealing with signed sums.
I fail to see why this is true.
 A: If a series is not absolutely convergent, then it only converges because some positive and negative terms cancel in the right way as you continue summing. However, upon rearrangement, you may find that the positive and negative terms do not cancel the same way. Consider the series
$$ 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} - \frac{1}{8} + \ldots$$
You can see that this sum does not converge because the partial sums oscillate between $0$ and $1$. However, let us rearrange the terms in the following manner:
$$ 1 - \frac{1}{2} + \left(- \frac{1}{2} + \frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{8}- \frac{1}{8}\right) + \left( \frac{1}{4} - \frac{1}{8} - \frac{1}{8}\right) + \left( - \frac{1}{8} + \frac{1}{16} + \frac{1}{16} \right) - \ldots$$
If you check this you will find that the terms of this series are the same as the terms of the previous series, but in a different order. However, this second series does converge.
What we can see here, intuitively, is that in the first series, too many positive and negative terms are grouped together, so they cancel each other out in "huge chunks", causing the oscillation between $0$ and $1$. However, in the second series, the positive and negative terms are interleaved in such a way that they cancel each other out in small "small chunks", causing oscillations that exponentially decrease in size.
