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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. Prove that $\sum\limits_{n=0}^\infty a_n=\sum\limits_{n=0}^\infty a_{\sigma(n)}$.

Proof:

We know that $\sum\limits_{n=0}^\infty a_{\sigma(n)}$ also converges absolutely. Consider a partial sum $S_N:=\sum\limits_{n=0}^N (a_n-a_{\sigma(n)})$. There exist $j_1, i_1 \in\mathbb{N}$ such that $j_1 \ge N$ and $j_1 \ge i_1$ and $S_{j_1}=\sum\limits_{n=0}^{j_1} (a_n-a_{\sigma(n)})=\underbrace{(a_1-a_{\sigma(i_1)})}_\text{=0}+...+...$ (after rearranging the terms). Similarly, there exist $j_2, i_2\in\mathbb{N}$ such that $j_2 \ge j_1$, $j_2\ge i_2$, and $S_{j_1}=\sum\limits_{n=0}^{j_1} (a_n-a_{\sigma(n)})=\underbrace{(a_1-a_{\sigma(i_1)})+(a_2-a_{\sigma(i_2)})}_\text{=0}+...+...$ Continuing in this fashion, we can conclude that there exist $j_M,i_M\in\mathbb{N}$ such that $j_M \ge j_{M-1}$ and $i_M\le j_M$, so that $S_{j_M}=\sum\limits_{n=0}^{j_M} (a_n-a_{\sigma(n)})=\underbrace{(a_1-a_{\sigma(i_1)})+...+ (a_M-a_{\sigma(i_M)})}_\text{=0} +...$

Hence, $\lim\limits_{M\to\infty} S_{j_M} = 0$.

Do you think this is rigorous enough? Also, I think that we don't need absolute convergence for this proof, do we?

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    we absolutely need absolute convergence. For a conditionally convergent series, a permutation can be found to make the sum take any value. – 2017-02-16
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    You can't informally rearrange the terms in a theorem about whether or not we can rearrange the terms. – 2017-02-16
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    @spaceisdarkgreen Do you mean that I can rearrange the terms only because the series is absolutely convergent? And thus my proof is correct? – 2017-02-16
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    Sure you could use the fact that an absolutely convergent series can be rearranged to undo the permutation and cancel off the terms pairwise, but that would be assuming the theorem that you're supposed to prove. – 2017-02-16

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Let $\epsilon >0.$ Let $$s_n= \sum_{k=1}^na_k$$ be the partial sum. By convergence $s_n\to s,$ we can pick an $N$ such that $|s_n-s| < \epsilon/2$ for $n>N.$ By absolute convergence, we have $$ \lim_{n\to \infty}\sum_{k=n}^\infty |a_k| = 0.$$ If need be, revise $N$ so that it's large enough so that we also have $$ \sum_{k=n}^\infty |a_k| <\epsilon/2$$ for all $n>N.$

Now, consider the permuted series $$ s'_n = \sum_{k=1}^na_{\sigma(k)}.$$ There is an $N'$ such that all of the elements $\{1\ldots N\}$ are included in $\{\sigma(1),\ldots,\sigma(N')\}.$ Then for $n>N',$ we have $$s'_{n} = s_N + r_n$$ where $r_n$ are what's left over after you gather $a_1+a_2+\ldots+a_N = s_N.$ Now, since all elements $a_1,\ldots,a_N$ are already in $S_N,$ all of the terms that make up $r_n$ are $a_i$ for $i>N.$ Thus we have $$ |r_n| \le \sum_{k=N+1}^\infty|a_k| \le \epsilon/2.$$

Finally we can put it together. For $n>N',$ $$|s_n'-s| = |s_N+r_n'-s| \le |s_n-s| + |r_n| \le \epsilon/2+\epsilon/2 =\epsilon. $$

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    I'm still stuck. For $n>N'>N$, $\| S'_{n} - S_{n} \| = \| S_n + R_n - S_n \| = \| R_n \|$. How can I show that $\| R_n \|\to 0$? – 2017-02-16
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    @sequence $\sum_{i=N}^\infty |a_i| \to 0$ as $N\to \infty.$ – 2017-02-16
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    Do you mean $|R_n| < \sum_{i=N'+1}^\infty |s_i'|$? – 2017-02-16
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    @sequence No, no prime. But I mistakenly said $s_i$ rather than $a_i$ – 2017-02-16
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    But it's $s'_n$ which contains additional terms for $n>N'>N$, not $s_n$. – 2017-02-16
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    @sequence The additional terms are all of the form $a_i$ for $i > N$ – 2017-02-16
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    So I think that for $n>N'>N$, it should be $R_n=\sum_{k=N+1}^\infty a_{\sigma(k)}=\lim\limits_{M\to\infty}\sum_{k=N+1}^M a_{\sigma(k)}$. But now take $W=\max\{\sigma(N+1), ..., \sigma(M)\}$, then $\sum_{k=N+1}^M a_{\sigma(k)}\le \sum_{k=N+1}^W a_k$. – 2017-02-16
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    @sequence I meant $R_n < \sum_{k=N+1}^\infty |a_{k}|$. This is just cause all the terms in the remainder have index greater than $N$. I converted the sketch to a (hopefully fully correct and rigorous) proof. – 2017-02-16
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53727/discussion-between-spaceisdarkgreen-and-sequence). – 2017-02-16
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    Very cool proof :) – 2017-02-17