Reorder the terms of the series This problem comes from a Brazilian book of Real Analysis. 
Explicitly perform a reordering of the terms of the series $$1 -\frac{1}{2} + \frac{1}{3} -\frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \cdots $$  so that its sum becomes zero.
It suggests to use the following reordering:
$$1 - \frac{1}{2} - \frac{1}{4} -  \frac{1}{6} - \frac{1}{8} + \frac{1}{3} - \frac{1}{10} - \frac{1}{12} - \frac{1}{14} - \frac{1}{16} + \frac{1}{5} - \frac{1}{18} - \frac{1}{20} - \frac{1}{22}-  \frac{1}{24} + \cdots$$
Where after each positive term is placed the first 4 negatives not used yet. But I don't know how to prove that this arrangement of the series converge to zero.

Its not actually four terms. Perhaps, to make the hint more obvious:
$$\begin{align}0&<1\\0&>1-\frac12-\frac14-\frac16-\frac18\\0&<1-\frac12-\frac14-\frac16-\frac18+\frac13\\0&>1-\frac12-\frac14-\frac16-\frac18+\frac13-\frac1{10}-\frac1{12}-\frac1{14}-\frac1{16}\end{align}$$
If you rearrange it so that the partial sums always go above zero and then below zero, and forever alternating in this manner, what should the limit be?
And can you prove that you can always pull the partial sums to the other side of zero? Hint: The harmonic series diverges.
