# Proof

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 $$( * * )$$:

$\begin{array}{rcl} \left|\sum_{k=0}^{m} x_{\sigma(k)}~-~L\right|&=&\left|\sum_{k=0}^{m} x_{\sigma(k)}~-~\sum_{k=0}^{N(\epsilon) - 1} x_{k}~+~\sum_{k=0}^{N(\epsilon) - 1} x_{k}~-~L\right|\\ &\le & \left|\sum_{k=0}^{m} x_{\sigma(k)}~-~\sum_{k=0}^{N(\epsilon) - 1} x_{k}\right|~+~\left|\sum_{k=0}^{N(\epsilon) - 1} x_{k}~-~L\right|\\ &\le& \sum_{k=N(\epsilon)}^{\infty} |x_{k}|~+~\frac\epsilon2\\ &\le& \frac\epsilon2~+~\frac\epsilon2=\epsilon. \end{array}$

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983