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.

Showing Convergence
Breaking the series into chunks of $3$ terms, which is okay since the terms tend to $0$, we get
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\left(\frac1{4k+1}-\frac1{4k+4}\right)+\left(\frac1{4k+3}-\frac1{4k+4}\right)\right]\\
&=\sum_{k=0}^\infty\left[\frac3{(4k+1)(4k+4)}+\frac1{(4k+3)(4k+4)}\right]
\end{align}
$$
Which can be compared to
$$
\left[\frac34+\frac3{16}\sum_{k=1}^\infty\frac1{k^2}\right]
+\left[\frac1{12}+\frac1{16}\sum_{k=1}^\infty\frac1{k^2}\right]
=\frac56+\frac14\sum_{k=1}^\infty\frac1{k^2}
$$
which converges by the $p$-test.

One Approach to Evaluation
If curious about the actual sum, it can be computed, using $(11)$ from this answer, as
$$
\begin{align}
&\frac14\sum_{k=1}^\infty\left[-\left(\frac1k-\frac1{k-\frac34}\right)-\left(\frac1k-\frac1{k-\frac14}\right)\right]\\
&=\frac14\left[-H_{-3/4}-H_{-1/4}\right]\\
&=\frac14\left[-(-\pi/2-3\log(2))-(\pi/2-3\log(2))\right]\\
&=\frac32\log(2)
\end{align}
$$

Another Approach to Evaluation
Using the fact that the alternating Harmonic Series converges to $\log(2)$, we get
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\frac1{4k+1}-\frac1{4k+2}+\frac1{4k+3}-\frac1{4k+4}\right]\\
&+\sum_{k=0}^\infty\left[\hphantom{\frac1{4k+1}}+\frac1{4k+2}\hphantom{\ \!+\frac1{4k+3}}-\frac1{4k+4}\right]\\
&=\log(2)+\frac12\log(2)\\[6pt]
&=\frac32\log(2)
\end{align}
$$
