I need to prove the following property in general but just don't know how to do it:
Let $\sum_{n=0}^\infty x_n$ be a series which converges conditionally in $\mathbb{C}$, so it doen't converge absolute. Show that there exists a bijection $\pi : \mathbb{N} \to \mathbb{N}$ so that the series $\sum_{n=0}^\infty x_{\pi(n)}$ does not converge.
I know it has something to do with rearrangements of the terms but I don't see how that can solve the problem. I know that if $\sum_{n=0}^\infty x_n$ converges absolute that after the bijection the series $\sum_{n=0}^\infty x_{\pi(n)}$ also converges, but I don't know if this is useful here.
Thanks in advance
Oops, I just noticed that you said $\mathbb{C}$. However, you can use Martin R's comment that $\sum x_n$ is absolutely convergent iff both $\sum \Re(x_n)$ and $\sum \Im(x_n)$ are absolutely convergent, and proceed as below.
Basic approach. The series is not absolutely convergent, meaning that the series $\sum y_n$ (with $y_n = |x_n|$) is not convergent. So divide $x_n$ into its positive terms and its negative terms. One of the two of those must not converge, because if each of them did converge, the series would have been absolutely convergent.
Thus, put whichever of the two subsets is not convergent first, and that ordering is the one you want.