Here is something wild: not only do rearrangements of conditionally convergent series give different answers, we can rearrange them to make them equal anything we want!
Let $\sum a_n$ be a conditionally convergent series, and define two new sequences:
$$a^+_n=\frac{a_n+|a_n|}{2}=\begin{cases}a_n&a_n\ge0\\0&a_n<0\end{cases}\\a^-_n=\frac{a_n-|a_n|}{2}=\begin{cases}a_n&a_n\le0\\0&a_n>0\end{cases}$$
By the divergence of $\sum|a_n|$, it follows that both $\sum a^+_n$ and $\sum a_n^-$ diverge, but we have $\lim_{n\to\infty}a_n^+=\lim_{n\to\infty}a_n^-=0$. Now, choose any $r\in\mathbb{R}$. We assume $r\ge 0$, but the proof is similar in the case $r\le 0$.
Without being too explicit, here is how to rearrange the terms of $\sum a_n$ so that the sum of the rearrangement is equal to $r$. Take the first terms to be the first terms of $a_n^+$ until the partial sum is greater than $r$. Then, allow the next terms to come from $a_n^-$ until the partial sum is less than $r$. Continue this process to obtain the desired rearrangement.
Two things we must note. First, the divergence of the positive and negative sums guarantees we will always be able to jump back and forth across $r$. Second, since the limit of the sequences go to zero, the jumping back and forth across $r$ will eventually get more and more precise, so that the limit will approach $r$.
Fascinating.
Let $b_n = (|a_n| + a_n)/2$ and $c_n = (|a_n|-a_n)/2.$
Then the partial sums satisfy
$$\sum_{n=1}^m a_n = \sum_{n=1}^m b_n - \sum_{n=1}^m c_n, \\ \sum_{n=1}^m |a_n| = \sum_{n=1}^m b_n + \sum_{n=1}^m c_n.$$
If $\sum a_n$ converges and $\sum |a_n|$ diverges, then both $\sum b_n$ and $\sum c_n$ diverge, since
$$2\sum_{n=1}^m b_n = \sum_{n=1}^m |a_n| + \sum_{n=1}^m a_n, \\ 2\sum_{n=1}^m c_n = \sum_{n=1}^m |a_n| - \sum_{n=1}^m a_n,$$
and the sum or difference of a divergent and convergent series is divergent.
Furthermore, we have divergence to $+\infty$ in each case, as the partial sums of $|a_n|$ form a non-negative, non-decreasing sequence.
Note that
$$\{b_n: n \in \mathbb{N}, b_n \neq 0\} = \{a_n: n \in A^+, a_n \neq 0\}, \\ \{c_n: n \in \mathbb{N}, c_n \neq 0\} = \{-a_n: n \in A^-\}, $$
and it easily shown that
$$ \sum_{n\in A_+} a_n=\sum_{n=1}^\infty b_n = +\infty\\ \sum_{n\in A_-} a_n = -\sum_{n=1}^\infty c_n = - \infty
$$
Best Answer
If the space is not complete, absolute convergence actually doesn't imply convergence. Take for example $X = \mathbb{Q}$ (with the norm inherited from $\mathbb{R}$) and choose a sequence of rational numbers $a_n$ such that $$ \sum_{n=1}^\infty a_n $$ is irrational, but $$ \sum_{n=1}^\infty |a_n| = 1. $$
See this question for details. Thus, in $X$, the series is absolutely convergent, but not convergent.