If $(a\_{n})$ is a conditionally convergent series in real field, then for any real number $\alpha$, there exists a rearrangement $(a\_{k\_{n}})$ of $(a\_{n})$ such that for all even $n$, $a\_{k\_{n}} \geq 0$, for all odd $n$, $a\_{k\_{n}} \leq 0$, and $\sum a\_{k\_{n}} = \alpha$.


This problem boils down to the following problem: can every conditional real series be rearranged to an alternative convergent series. If this is solved, then apply the extensions of Riemann's theorem by [Sierpiński](https://mathoverflow.net/questions/47589/how-to-rearrange-only-negative-part-of-a-conditionally-convergent-series-to-get), the original problem is done.


I appreciate any suggestion about this problem. Thanks in advance.



A restricted version of Riemann series theorem: rearrangements with alternating signs
Here you prescribe in addition the sequence of signs of the rearranged series in the Riemann-Dini theorem to be alternating, but note that any non-stationary binary sequence of signs does as well. More precisely:



> 
> Let $(a\_k)\_{k\in\mathbb N} $ be an infinitesimal sequence of non-zero real
> numbers such that $\sum\_{k\ge0}a^+\_k=\sum\_{k\ge0}a^-\_k=+\infty$.
> 
> 
> Let $\epsilon\in \{-1,1\}^\mathbb{N}$ be a non-eventually constant
> sequence.
> 
> 
> Let $\alpha\in\mathbb R \cup\{ \pm\infty\}$.
> 
> 
> Then there exists a permutation $\sigma$ of $\mathbb N$ such that
> $$\sum\_{k=0}^\infty a\_{\sigma(k)}=\alpha$$
> $$\text{sgn}\,a\_{\sigma(k)}=\epsilon\_k.$$
> 
> 
> 


To this end: extract a subset $S\subset\mathbb N$ such that $\sum\_{k\in S}a\_k$ is absolutely summable and $a\_k$ are positive resp. negative for infinitely many $k\in S$ (therefore for infinitely many $k\in \mathbb N\setminus S$ as well, because of the assumption $\sum\_{k\ge0}a^+\_k=\sum\_{k\ge0}a^-\_k=+\infty$).


Then do the Riemann-Dini bijection $\tau:\mathbb N\to \mathbb N \setminus S$ relatively to the sequence $(a\_k)\_{k\in \mathbb N\setminus S }$ and the number $\alpha-\sum\_{k\in S}a\_k$, namely $$\sum\_{k=0}^\infty a\_{\tau(k)}=\alpha-\sum\_{k\in S}a\_k.$$


Finally, insert the coefficients $\{a\_k\}\_{k\in S}$ in some order into the series $\sum\_{k=0}^\infty a\_{\tau(k)}$, so as to get a rearrangement with the prescribed final sequence of signs $(\dagger)$. Since $\sum\_{k\in S}a\_k$ is absolutely summable, the order of the insertion does not affect the convergence and the value of the sum, which is $\alpha$ as wanted.


$(\dagger)$ This can easily be done, for the reason that any non eventually constant binary sequence contains any other non eventually constant binary sequence as a subsequence, in such a way that the complement is also non eventually constant.


