Baby Rudin claim: $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}...$ converges This sequence is a rearrangement of the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}...$. Note that at this point in the text we do not have any theorem about the convergence of rearrangements.
Let $\{s_n\}$ be the sequence of partials sums of the series then for $n \ge 0$
$s_{3(n+1)} = \sum ^n _ {k=0} \frac{1}{4k+1} + \frac{1}{4k+3} - \frac{2}{4k+4}$
We can view it as the sequence(on $n$) of partials sums of
$\sum_0 a_n = \sum_0 \frac{1}{4n+1} + \frac{1}{4n+3} - \frac{2}{4n+4}$
Where $|a_n| = a_n = \frac{1}{4n+4}\{\frac{3}{4n+1}+\frac{1}{4n+3}\} \le \frac{1}{4n^2}$.
By the comparison test $s_{3(n+1)}$ converges to some real $\alpha$.
But $s_{3(n+1)+1} = s_{3(n+1)}+ \frac{1}{4n+5}$ and $s_{3(n+1)+2} = s_{3(n+1)}+ \frac{1}{4n+5}+\frac{1}{4n+7} $hence we have a partition of $\{s_n\}$ into subsequences which tend to $\alpha$ and this implies $s_n \rightarrow \alpha$.
Is my proof correct? Any alternative solutions are appreciated.

The following argument shows that the series converges, and gives its sum:
$\hspace{.3 in}1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots=\ln 2$ $\;\;\;$so
$\hspace{.27 in}\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots=\frac{1}{2}\ln 2$.  $\;\;\;$Inserting zeros, we get
$\hspace{.26 in}0+\frac{1}{2}+0-\frac{1}{4}+0+\frac{1}{6}+0-\frac{1}{8}+0+\frac{1}{10}+\cdots=\frac{1}{2}\ln 2$.  
Adding this to the original series gives
$\hspace{.26 in}1+0+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+0+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+0+\cdots=\frac{3}{2}\ln 2$, $\;\;$ so
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.
