Rearranging a series. Consider the following two series.
$$\sum a_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+ \cdots$$ $$\sum b_n = 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+ \cdots$$
I was able to say that $a_n = (-1)^{n+1} \frac{1}{n}$ while $b_n = \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}$, which was obtained after rearranging the numbers.
I understand up to the fact that if a series is not absolutely convergent, rearranging the sequence will change the limit of the sum.
I haven't seen the proof yet, but I have two questions.
1), I want to prove that the second series converges. The ratio test was inconclusive, so trying to use the root test I got the following. Can someone check my work ? Is this a natural approach ?

$$\sqrt[n]{b_n} =\sqrt[n] {\frac{8n-3}{2n(4n-1)(4n-3)}} \le \sqrt[n]{\frac{8}{32n^2}}$$
so $$\lim_{n \to \infty}\sqrt[n]{b_n}=1$$
which is inconclusive, again.

2), Assuming that convergence was known, how can we calculate the limit of the sum ?
 A: Since 
$$
\small
b_n=\frac{8n-3}{2n(4n-1)(2n-3)}\sim\frac{1}{2n^2}
$$
The series $\sum b_n$ absolutely converges. To find its sum we need to work a little. Note that
$$
\small
\begin{align}
\sum\limits_{k=1}^n b_k
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)+\color{green}{\left(\frac{1}{2n}+\frac{1}{2n+1}+\frac{1}{2n+3}+\ldots+\frac{1}{4n-1}\right)}\\
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)\color{green}{+\frac{1}{2n}}\color{green}{+\frac{1}{2n+1}}\color{red}{+\frac{1}{2n+2}}\color{green}{+\frac{1}{2n+3}}\color{red}{+\frac{1}{2n+4}}\color{green}{+\ldots+\frac{1}{4n-1}}\color{red}{+\frac{1}{4n}}\color{red}{-\left(\frac{1}{2n+2}+\frac{1}{2n+4}+\ldots+\frac{1}{4n}\right)}\\
&=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots-\frac{1}{4n}\right)\color{green}{+\left(\frac{1}{2n}+\ldots+\frac{1}{4n}\right)}\color{red}{-\frac{1}{2}\left(\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}\right)}\\
\end{align}
$$
Thus
$$
\small
\begin{align}
\sum\limits_{k=1}^\infty b_k
&=\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{4n}\frac{(-1)^{k-1}}{k}+\sum\limits_{k=2n}^{4n}\frac{1}{k}-\frac{1}{2}\sum\limits_{k=n+1}^{2n}\frac{1}{k}\right)\\
&=\lim\limits_{n\to\infty}\sum\limits_{k=1}^{4n}\frac{(-1)^{k-1}}{k}+\lim\limits_{n\to\infty}\left(\ln\frac{4n}{2n}+o(n)-\frac{1}{2}\left(\ln\frac{2n}{n+1}+o(n)\right)\right)\\
&=\ln 2+\ln 2-\frac{1}{2}\ln 2=\frac{1}{2}\ln 8
\end{align}
$$
