<s>[INST] 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.
 [/INST] 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.
</s>