(related to Proposition: Rearrangement of Absolutely Convergent Series)
From the definition of absolute convergence it follows that \(\sum_{k=0}^\infty |x_k|\) is convergent to a limit, say \(L\). We have to show that for every permutation \(\sigma:\mathbb N\to \mathbb N\), the rearrangement \(\sum_{k=0}^\infty x_{\sigma(k)}\) is convergent to \(L\).
Assume \(\epsilon > 0\). Because \(\sum_{k=0}^\infty |x_k|=L\), and applying the triangle inequality, there exists an index \(N(\epsilon)\) such that
\[\left|\sum_{k=0}^{N(\epsilon) - 1} x_{k}~- L\right|=\left|\sum_{k=N(\epsilon)}^{\infty} x_{k}\right|\le \sum_{k=N(\epsilon)}^{\infty} |x_{k}| < \frac\epsilon2.\quad\quad ( * )\]
Consider the subset of natural numbers given by
\[\Sigma(M):=\{\sigma(0), \sigma(1), \sigma(2), \ldots, \sigma(M)\}.\]
Note that we can choose \(M\) big enough to satisfy \[\{0, 1, 2, \ldots, N(\epsilon)-1 \}\subset \Sigma(M). \]
This enforces the inequality
\[\left|\sum_{k=0}^{m} x_{\sigma(k)}~-~\sum_{k=0}^{N(\epsilon) - 1} x_{k}\right|\le \sum_{k=N(\epsilon)}^{\infty} x_{k}\le \sum_{k=N(\epsilon)}^{\infty} |x_{k}|,\quad\quad ( * * )\]
since all sequence members \(x_k\), \(k=0,1,\ldots,N(\epsilon) - 1\) cancel out, leaving only sequence members \(x_k\) left for \(k \ge N(\epsilon)\). For all \(m\ge M\), it follows from \(( * ) \) and \(( * * ) \):
\[\begin{array}{rcl} \left|\sum_{k=0}^{m} x_{\sigma(k)}~-~L\right|&=&\left|\sum_{k=0}^{m} x_{\sigma(k)}~-~\sum_{k=0}^{N(\epsilon) - 1} x_{k}~+~\sum_{k=0}^{N(\epsilon) - 1} x_{k}~-~L\right|\\ &\le & \left|\sum_{k=0}^{m} x_{\sigma(k)}~-~\sum_{k=0}^{N(\epsilon) - 1} x_{k}\right|~+~\left|\sum_{k=0}^{N(\epsilon) - 1} x_{k}~-~L\right|\\ &\le& \sum_{k=N(\epsilon)}^{\infty} |x_{k}|~+~\frac\epsilon2\\ &\le& \frac\epsilon2~+~\frac\epsilon2=\epsilon. \end{array} \]