Any rearrangement of an absolutely convergent sequence $(a_n)$ is another absolutely convergent sequence with the same limit. Let $(a_{\sigma(n)})$ be the rearranged sequence under the bijection of indices $\sigma$.
My proof attempt. Let $T_N = \sum_{n=1}^N |a_{\sigma(n)}|$ be the $N$th partial sum of the rearranged absolute series. Similarly let $S_N$ be the partial sum of the original absolute series. I know I want to try to prove the absolute sequences Cauchy. Still working on it.
Best Answer
You needn't make the $N$s match. You know you can take $N$ large enough so that $$\sum_{k\geqslant N}|a_k|<\varepsilon$$
Now pick $M$ large enough so that all the terms in $T_M$ are some of the terms in $S_N$. Then how large is $$\left|\sum_{k\geqslant 0}a_k-T_M\right|?$$