Generalization of the Riemann rearrangement Theorem

Let $(X, \| \cdot \|)$ be a Banach space and $(x_n) \subset X$ such that $\sum_{n} \|x_n\| = +\infty$. Is it true that for each $x\in X$ there is a bijection $\phi :\mathbb{N}\to\mathbb{N}$ such that $\sum_{n} x_{\phi(n)} = x$ ? I believe that inductively this can be proven if $X$ is of finite dimension. But what about higher dimensions?

• You (at least) also have to assume $\displaystyle\sum x_n$ to be conditionnally convergent. But still, I don't think it holds : pick $X=\Bbb{R}^2$, pick your favourite conditionnaly convergent not absolutely convergent series $\displaystyle\sum x_n$, and your favourite absolutely convergent series $\displaystyle\sum y_n$. Then the sequence $(x_n, y_n)$ would be a counterexample because no matter how you rearrabge the terms, the sum of the second coordinate will always be $\displaystyle\sum_{n=0}^\infty y_n$ (if the limit makes sense) – Max Jun 25 '17 at 10:40
• Some references to literature related to some results on rearrangements of series in Banach spaces are mentioned in this answer. – Martin Sleziak Jun 25 '17 at 12:02

Suppose $X$ is just $\mathbb R^2$ with basis $\{x,y\}$.Take the standard conditionally convergent sequence of real numbers $a_n = (-1)^nn^{-1}$ and consider the sequence $v_n=(a_n,0)$ of vectors. By Riemann's theorem we can for any $r \in \mathbb R$ rearrange the sequence to make any $(r,0)$ the limit. But we cannot for example rearrange to make $(0,1)$ the limit.
• @Kps Look at $\frac{(-1)^n}n\, (1,1)$. If you rearrange you will never get anything that does not lie on the diagonal of $\Bbb R\times \Bbb R$. – s.harp Jun 25 '17 at 11:05
• Let $S\subset X^*$ countable, such that the span$(S)$ is dense in $X^*$. If $\sum x^*(x_n)$ converges, while $\sum |x^*(x_n)|=\infty$, for all $x^*\in X^*$, then for every $x\in X$, there is a rearrangement $x_{j_n}$ of $x_n$, such that $\sum x_{j_n}=x$. – Yiorgos S. Smyrlis Jun 25 '17 at 11:17