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.
 A: By the Riemann-Dini theorem, we may take any series that is conditionally convergent but not absolutely convergent and rearrange it in order to get a series that converges to $\alpha$, for any $\alpha\in\mathbb{R}$.
In our case:
$$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2}\right)&=&\sum_{k\geq 0}\int_{0}^{1}\left(x^{4k}+x^{4k+2}-2 x^{4k+3}\right)\,dx\\&=&\int_{0}^{1}\frac{1+x^2-2x^3}{1-x^4}\,dx\\&=&\frac{3}{2}\log 2.\end{eqnarray*} $$
We may notice that we know in advance that the LHS is converging, since:
$$ \frac{1}{4k+1}+\frac{1}{4k+3}-\frac{1}{2k+2} = \frac{8k+5}{(4k+1)(4k+3)(2k+2)}=O\left(\frac{1}{k^2}\right).$$
Convergence also follows from Dirichlet's test, since the sequence $1,1,-2,1,1,-2,\ldots$ has bounded partial sums while the sequence $\frac{1}{1},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{7},\frac{1}{8},\ldots$ decreases to zero.
A: 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
$$
\begin{align}
\sum_{k=0}^\infty\left[\frac1{4k+1}+\frac1{4k+3}-\frac2{4k+4}\right]
&=\sum_{k=0}^\infty\left[\frac1{4k+1}-\frac1{4k+2}+\frac1{4k+3}-\frac1{4k+4}\right]\\
&+\sum_{k=0}^\infty\left[\hphantom{\frac1{4k+1}}+\frac1{4k+2}\hphantom{\ \!+\frac1{4k+3}}-\frac1{4k+4}\right]\\
&=\log(2)+\frac12\log(2)\\[6pt]
&=\frac32\log(2)
\end{align}
$$
A: 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
$\hspace{.25 in} 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\cdots=\frac{3}{2}\ln 2$.
A: 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 $s=\frac{3}{2}\ln(2)$
