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Let $\sum\limits_{n=0}^\infty a_n$ be an absolutely convergent series in $\mathbb{R}^n$ over the Eucledian norm, and $\sigma:\mathbb{N}_0\to \mathbb{N}_0$ (where $\mathbb{N}_0=\mathbb{N}\cup \{0\})$ be a permutation. Then $\sum\limits_{n=0}^\infty a_{\sigma(n)}$ also converges absolutely.

Proof:

$\sigma(n)$ simply changes the order of the summation terms. So $\infty>\sum\limits_{n=0}^\infty \|a_n\|=\sum\limits_{n=0}^\infty \|a_{\sigma(\sigma^{-1}(n))}\|=\lim\limits_{N\to\infty}\sum\limits_{k=\sigma^{-1}(0)}^{\sigma^{-1}(N)} \|a_{\sigma(k)}\|=\lim\limits_{N\to\infty}\sum\limits_{k=0}^{\sigma^{-1}(N)-\sigma^{-1}(0)} \|a_{\sigma(k+\sigma^{-1}(0))}\|=\lim\limits_{N\to\infty}\sum\limits_{k=0}^{\sigma^{-1}(N)-\sigma^{-1}(0)} \|a_{\sigma(k)}+a_{\sigma^{-1}(0))}\|\ge \lim\limits_{N\to\infty}\left\|\sum\limits_{k=0}^{\sigma^{-1}(N)-\sigma^{-1}(0)} \left(a_{\sigma(k)}+a_{\sigma^{-1}(0))}\right)\right\|=\lim\limits_{N\to\infty}\left\|\sum\limits_{k=0}^{\sigma^{-1}(N)-\sigma^{-1}(0)} a_{\sigma(k)}+(\sigma^{-1}(N)-\sigma^{-1}(0))a_{\sigma^{-1}(0))}\right\|$

Oops, I think I didn't choose the right way, but I hoped this could lead me to the desired result.

I guess it would be much more intuitive to build the proof like this:

$\lim\limits_{N\to \infty}\sum\limits_{n=0}^N \|a_{\sigma(n)}\|=\lim\limits_{N\to \infty} (\|a_{\sigma(0)}\|+...+\|a_{\sigma(N)}\|)$. Properly rearranging the terms, we can get $\lim\limits_{N\to \infty} (\|a_{\sigma(0)}\|+...+\|a_{\sigma(N)}\|)=\lim\limits_{N\to \infty} (\|a_{m_0}\|+...+\|a_{m_N}\|)$ (since a permutation on $N$ terms will give $N$ terms). We can find some $K>N$ such that $ (\|a_{m_0}\|+...+\|a_{m_N}\|)= (\|a_{0}\|+...+\|a_{m_K}\|)$. Constructing the series in this way, we obtain $\lim\limits_{K\to \infty}(\|a_{0}\|+...+\|a_{m_K}\|)=\sum\limits_{n=0}^\infty \|a_{n}\|$.

But this is too "heuristic". Should I use sequences of partial sums instead, or is there a way to continue my line of reasoning with the first chain of inequalities? Would appreciate your suggestion. The concept is obvious, but how to formalize the proof it seems is not a trivial task.

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    Let $m(N) = \max \{ \sigma(0), \sigma(1),\dotsc, \sigma(N)\}$. Then $$\sum_{n = 0}^N \lVert a_{\sigma(n)}\rVert \leqslant \sum_{n = 0}^{m(N)} \lVert a_n\rVert.$$ – 2017-02-15
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    @DanielFischer awesome! – 2017-02-16

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I think you can just argue this way: denote the $nth$ partial sum of the series of absolute values by $s_n$ and let $l$ be the limit. Fix $\epsilon >0.$ There is an integer $N$ such that $n>N\Rightarrow |s_n-l|<\epsilon/2.$ There is also an integer $J$ such that $\left \{ 1,\cdots, N \right \}\subseteq \left \{ \sigma(1),\cdots , \sigma(J)\right \}.$ Then, with $s'_j=\sum_{i=1}^{j}|a_{\sigma(i))}|,\ $ if $j>J,\ |s'_j-l|\le |s'_j-s_N|+|s_N-l|<\epsilon/2+\epsilon/2=\epsilon,\ $so $s'_j$ converges to $l$.