## Warmup (you've probably seen this before)

Suppose $\sum_{n\ge 1} a_n$ is a conditionally convergent series of real numbers, then by rearranging the terms, you can make "the same series" converge to any real number $x$. To do this, let $P=\{n\ge 1\mid a_n\ge 0\}$ and $N=\{n\ge 1\mid a_n<0\}$. Since $\sum_{n\ge 1} a_n$ converges conditionally, each of $\sum_{n\in P}a_n$ and $\sum_{n\in N}a_n$ diverge and $\lim a_n=0$.

Starting with the empty sum (namely zero), build the rearrangement inductively. Suppose $\sum_{i=1}^m a_{n_i}=x_m$ is the (inductively constructed) $m$-th partial sum of the rearrangement. If $x_m\le x$, take $n_{m+1}$ to be the smallest element of $P$ which hasn't already been used. If $x_m> x$, take $n_{m+1}$ to be the smallest element of $N$ which hasn't already been used.

Since $\sum_{n\in P}a_n$ diverges, there will be infinitely many $m$ for which $x_m\ge x$, so $n_{m+1}$ will be in $N$ infinitely often. Similarly, $n_{m+1}$ will be in $P$ infinitely often, so we've really constructed a rearrangement of the original series. Note that $|x-x_m|\le \max\{|a_n|\bigm| n\not\in\{n_1,\dots, n_m\}\}$, so $\lim x_m=x$ because $\lim a_n=0$.

Suppose $\sum_{n\ge 1}v_n$ is a conditionally convergent series with $v_n\in \mathbb R^k$. Can the sum be rearranged to converge to any given $w\in \mathbb R^k$?

Obviously not! If $\lambda$ is a linear functional on $\mathbb R^k$ such that $\sum \lambda(v_n)$ converges absolutely, then $\lambda$ applied to any rearrangement will be equal to $\sum \lambda(v_n)$. So let's also suppose that $\sum \lambda(v_n)$ is conditionally convergent for every non-zero linear functional $\lambda$. Under this additional hypothesis, I'm pretty sure the answer should be "yes".