Why commutatively convergent series iff summable? maybe this is a stupid question, however I could not figure out its solution. I´am assuming that a summable series (convergent to $x$) is a series in a normed space $E$ indexed by $I$ such that, for each $\varepsilon > 0$, there exists some $F_{\varepsilon} \subset I$ such that, for any $F_{\varepsilon} \subset F$, $||\sum_{i \in F} x_i - x|| < \varepsilon$ holds.
I´ve seen in a book that, in $\mathbb{R}$ and $\mathbb{C}$, summable and comutatively convergent are equal, however I could not prove that commutatively convergent implies  summable (I cannot find a global $N$ such that the sum converges for any permutation). Furthermore, I want to know if this equivalence holds for arbitrary normed spaces (or Banach spaces), if not, is there a nice counter-example?
Thanks in advance. 
 A: This is true in the Banach space setting.
Argue by contraposition. 
Choose $\epsilon_0$ such that for any finite set $F$, there is a finite set $F'\supset F$ with 
$$\biggl\Vert x-\sum_{i\in F'} x_i\biggr\Vert\ge\epsilon_0.$$ 
Choose $N$ so that for $n\ge N$, 
$$\biggl\Vert x-\sum_{i=1}^n x_i\biggr\Vert\le\epsilon_0/2.$$  Now let $F_1=\{1,\ldots,N\}$ and choose $F_1' \supset F_1$ so that 
$$\biggl\Vert x-\sum_{i\in F_1'} x_i\biggr\Vert\ge\epsilon_0.$$ 
Let $F_2=\{1, \ldots, \max\limits_{i\in F_1'} i\}$ and choose $F_2'\supset F_2$ so that 
$$\biggl\Vert x-\sum_{i\in F_2'}  x_i\biggr\Vert\ge\epsilon_0 .$$
Construct $F_3, F_3', F_4, F_4', \ldots$ in the same way.
Define a permutation $\sigma$ of $\Bbb N$ by enumerating the elements of the sets
$$
F_1,\ F_1'\setminus F_1,\ F_2\setminus F_1',\ F_2'\setminus F_2,\ldots
$$
Then $\sum\limits_{n=1}^\infty x_{\sigma(n)}$ is not convergent.
Indeed, the series is not Cauchy: for any $n$
$$\eqalign{\biggl\Vert
\sum_{i\in F_n'\setminus F_n} x_i\biggr\Vert
&=\biggl\Vert (x-\sum_{i\in F_n} x_i )- (x-\sum_{i\in F_n'}x_i) \biggr \Vert\cr
&\ge\biggl|\,\biggl  \Vert (x-\sum_{i\in F_n} x_i )\biggr\Vert-\biggl\Vert (x-\sum_{i\in F_n'}x_i)  \biggr\Vert \,\biggr |\cr
&\ge \epsilon_0-\epsilon_0/2\cr&=\epsilon_0/2.
}$$

The above was taken from Lemma 16.1, page 459, of Bases in Banach Spaces I, Ivan Singer, 
A: The equivalence holds in any Banach space. You can find the proof in virtually any book on Banach space theory; for example, see Lemma 2.4.2 in the book "Topics in Banach space theory" by Albiac and Kalton. The idea is to proceed by contradiction: assuming that the series is not summable, it is possible to produce a permutation such that the partial sums of $\sum x_{\pi(i)}$ do not satisfy Cauchy's criterion.
In a finite-dimensional space, this also equivalent to absolute convergence, i.e. convergence of $\sum\Vert x_i\Vert$; but this is no longer true in any infinite-dimensional space, by the so-called Dvoretzky-Rogers theorem.
