Given a conditionally convergent series, can we group terms together retaining order, so that the quotient of consecutive groups tends to $1$? Suppose $\displaystyle\sum_n a_n$ is a conditionally convergent series, that is, $\displaystyle\sum_n a_n$ converges and $\displaystyle\sum_n \vert a_n \vert = +\infty.$
Proposition: There exists $\ k_1< k_2< k_3 < \ldots\ $ such that:
$$ \large{\frac{\sum_{n=k_i}^{n=k_{i+1}-1} a_n}{\sum_{n=k_{i+1}}^{n=k_{i+2}-1} a_n}}\ \overset {i\to\infty}{\longrightarrow}\ 1. $$
Edit: I think my original question intended to be for:
$$ \left\vert \large{\frac{\sum_{n=k_i}^{n=k_{i+1}-1} a_n}{\sum_{n=k_{i+1}}^{n=k_{i+2}-1} a_n}}\right\vert\ \overset {i\to\infty}{\longrightarrow}\ 1. $$
Since these question are (seemingly closely related), and someone has already put effort into answering the first question, I will leave both questions as part of this one question, if that makes sense.
Is this true? Further, can we replace $1$ here with any number and it still be true?
I thought this was interesting and was wondering if it was true or false...
Obviously this is false for absolutely convergent series, for example $a_n = 2^{-n}.$
 A: If $n \geq 0$ and $3^n \leq j < 3^{n+1}$
set $$a_{j} = \frac{(-1)^{j}}{3^{n}}.$$
It is clear that $\sum_{j=1}^{\infty} |a_{j}| = \infty.$
Set $m_{3}(u) = \inf\{3^{t}:t\in \mathbb{N}\cup\{0\}, 3^{t} > u\}$ and $M_{3}(u) = \max \{3^{t}: t \in \mathbb{N}\cup\{0\}, 3^{t} \leq u\}.$
For $r,s \in \mathbb{N}$ $\sum_{j=r}^{s}a_{j}$ evaluates to one of $$0, \frac{1}{m_{3}(r)}, -\frac{1}{m_{3}(r)}, \frac{1}{m_{3}(r)} -\frac{1}{M_{3}(s)}, -\frac{1}{M_{3}(s)}, \frac{1}{M_{3}(s)}.$$
Suppose that there exists $k_{1}<k_{2}<...$ according to your proposition. It is necessary for all sufficiently large $i$ that $$\sum_{k_{i}}^{k_{i+1}-1}a_{d} \neq 0$$
for all $i \in \mathbb{N}$. Without loss of generality, we can assume that for all $i \in \mathbb{N}$ we have the previous fact. Now note that for all $h \in \mathbb{N}$ sufficiently large there exists $s_{h} \in \mathbb{N}$ so that $k_{s_{h}+1}\leq 3^{h} \leq k_{s_{h}+2}-1$ so that we have
$$\left\vert\sum_{d=k_{s_{h}}}^{k_{s_{h}+1}-1}a_{d}\right\vert \geq \frac{1}{3^{h-1}}.$$
and $$\left\vert\sum_{d = k_{s_{h}+1}}^{k_{s_{h}+2}-1} a_{d}\right\vert \leq \frac{1}{3^{h}}.$$
In particular
$$\frac{\left\vert\sum_{d=k_{s_{h}}}^{k_{s_{h}+1}-1}a_{d}\right\vert}{\left\vert\sum_{d = k_{s_{h}+1}}^{k_{s_{h}+2}-1} a_{d}\right\vert} \geq 3$$
which disproves your proposition.
