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.

Here is a general result:
Let $s$ be an infinite sum over a sequence $(a_n)_{n\in\mathbb N}$, where $a_n$ are a permutation of $\frac{(-1)^{k+1}}{k}$, but where the sub sequences $(a_{n_k})_{k\in\mathbb N},\space a_{n_k}=-\frac{1}{2k}$ and $(a_{m_k})_{k\in\mathbb N},\space a_{m_k}=\frac{1}{2k-1}$ exist with $n_k<n_{k+1}$ and $m_k<m_{k+1}$. So
$$
s=1+\frac13+\frac15-\frac12+\frac17+\frac19+\frac1{11}-\frac{1}{4}+...
$$
is such a sum, but
$$
s=1-\frac14+\frac13-\frac12+...
$$
isn't.
Now, for such a sequence, define $p_n$ as the number of plus signs up to $a_n$, $q_n$ as the number of minus signs up to $a_n$ and $l=\lim_{n\to\infty}\frac{p_n}{q_n}$.
We have:
$$
\sum_{k=1}^{n} a_k=\sum_{k=1}^{p_n}\frac{1}{2k-1}-\sum_{k=1}^{q_n}\frac{1}{2k}
$$
But also:
$$
\sum_{k=1}^{p_n}\frac{1}{2k-1}=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k}
$$
And therefore, if $H_n$ is the $n$-th harmonic number and using the well known result $H_n=\ln(n)+\gamma+\epsilon_n$ where $\lim_{n\to\infty}\epsilon_n=0$, we obtain:
$$
\sum_{k=1}^{n} a_k=\sum_{k=1}^{2p_n}\frac{1}{k}-\sum_{k=1}^{p_n}\frac{1}{2k}-\sum_{k=1}^{q_n}\frac{1}{2k}=\\
\sum_{k=1}^{2p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{p_n}\frac{1}{k}-\frac{1}{2}\sum_{k=1}^{q_n}\frac{1}{k}=\\
H_{2p_n}-\frac{1}{2}H_{p_n}-\frac{1}{2}H_{q_n}=\\
\ln{(2p_n)}+\gamma+\epsilon_{2p_n}-\frac{1}{2}\ln{(p_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\ln{(q_n)}-\frac{1}{2}\gamma-\frac{1}{2}\epsilon_{q_n}=\\
\ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n}
$$
And thus:
$$
\lim_{n\to\infty}\sum_{k=1}^{n} a_k=\lim_{n\to\infty}\left(\ln{(2)}+\frac{1}{2}\ln{\left(\frac{p_n}{q_n}\right)}+\epsilon_{2p_n}-\frac{1}{2}\epsilon_{p_n}-\frac{1}{2}\epsilon_{q_n}\right)=\ln(2)+\frac{1}{2}\ln(l)
$$
In the original sum, we have $l=2$ and therefore $s=\frac{3}{2}\ln(2)$
