Reordering

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Proof: Let \(a=\sum_{k=1}^\infty a_k\) and let \(\tau:\mathbb{N}\to\mathbb{N}\) be bijective. Let \(\varepsilon>0\). Since we have absolute convergence, there exists some \(N_1\) such that \[\sum_{k=N_1}^\infty |a_k|<\frac\varepsilon2\] and thus \[\left|a-\sum_{k=1}^{N_1-1} a_k\right|=\left|\sum_{k=N_1}^{\infty} a_k\right|\leq\sum_{k=N_1}^{\infty} |a_k|<\frac\varepsilon2.\] Now choose \(N:=\max\{\tau^{-1}(1),\tau^{-1}(2),...,\tau^{-1}(N_1-1)\}\).Then \[\{1,2,\ldots,N_1-1\}\subset\{\tau(1),\tau(2),\ldots,\tau(N)\}.\] Then, for all \(n\geq N\) holds \[\left|a-\sum_{k=1}^{n} a_{\tau(k)}\right|\leq \left|a-\sum_{k=1}^{N_1-1} a_k\right|+\left|\sum_{k=1}^{N_1-1} a_k-\sum_{k=1}^{n} a_{\tau(k)}\right|< \frac\varepsilon2+\sum_{k=N_1}^{\infty} |a_k|<\frac\varepsilon2+\frac\varepsilon2=\varepsilon.\] The absolute convergence of \(\sum_{k=1}^\infty a_{\tau(k)}\) can be shown by an application of the above argumentation to the series \(\sum_{k=1}^\infty |a_k|\) and \(\sum_{k=1}^\infty |a_{\tau(k)}|\). \(\Box\)