Example 3.53 in Baby Rudin Here's Example 3.53 in the book Principles of Mathematical Analysis by Walter Rudin, third edition. 

Consider the convergent series $$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \ldots$$ and one of its rearrangements $$1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{5} + \frac{1}{7} - \frac{1}{4} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} + \ldots$$ in which two positive terms are always followed by one negative. If $s$ is the sum of the original series, then $$s < 1 - \frac{1}{2} + \frac{1}{3} = \frac{5}{6}.$$ Since $$\frac{1}{4k-3} + \frac{1}{4k-1} - \frac{1}{2k} = \frac{8k-4}{(4k-1)(4k-3)} - \frac{1}{2k} = \frac{2k(8k-4) - (4k-1)(4k-3)}{2k(4k-1)(4k-3)} = \frac{8k-3}{2k(4k-1)(4k-3)} > 0$$ for $k \geq 1$, we see that $$s^\prime_3 < s^\prime_6 < s^\prime_9 < \ldots,$$ where $s^\prime_n$ is the $n$th partial sum of the series after the rearrangement. Hence $$\lim_{n\to\infty}\sup s^\prime_n > s^\prime_3 = \frac{5}{6},$$ so that the rearranged series certainly does not converge to $s$. 

Now here's my question:

How to determine, using the machinery developed by Rudin upto this point in the book, if the new (or rearranged) series converges at all? Rudin leaves it to the reader to check that the new series does converge. How to prove this convergence? 

I would like to have answers that use only the results that Rudin has discussed so far in the book. 

Hint: Notice that $s_{3k}'$ converge as $k \to \infty$. How far can $s_{3k+1}$ and $s_{3k+2}$ be from $s_{3k}$?
