Example to show that unconditional convergence does not imply absolute convergence in infinite-dimensional normed spaces. I tried to show that unconditional convergence does not imply absolute convergence in infinite-dimensional normed spaces using a direct proof, but unfortunately I did not succeed. The definition of unconditional convergence I am using is the following:

A series $\sum_n x_n$ is called unconditionally convergent if the series $\sum_n x_{\pi(n)}$ converges for all permutations $\pi : \mathbb{N} \to \mathbb{N}$.

I considered the sequence $(x_n)_{n \in \mathbb{N}} \subset c_0 (\mathbb{N})$, where $x_n = \frac{\delta_n}{n}$ and $\delta_n$ denoting the standard unit vector. It is clear that $\sum_n \| x_n \|_{\infty}$ does not convergence, but I have trouble showing that the series $\sum_n x_n$ converges unconditionally. 
Any help is appreciated. Thanks in advance.

It's pretty much straightforward.  Let $y$ be the limit: the member of $c_0$ with entries $y_j = 1/j$ for each $j$.  For any $\epsilon > 0$,  $\left\|\sum_{n=1}^N x_{\pi(n)} - y \right\| \le \epsilon$ if $\{\pi(1), \ldots, \pi(N)\}$ contains all positive integers $\le 1/\epsilon$.  For any permutation $\pi$, this must be the case for large enough $N$.
