Series can be reordered.

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**Definition 1** (Reordering). *Let \(\sum_{k=1}^\infty a_k\) be a series in
\(\mathbb{K}\) and let \(\tau:\mathbb{N}\to\mathbb{N}\) be
a bijective mapping. Then the series \[\sum_{k=1}^\infty a_{\tau(k)}\] is called
a **reordering* of \(\sum_{k=1}^\infty
a_k\).

**Theorem 2** (Reordering of absolutely convergent
series). *Let \(\sum_{k=1}^\infty
a_k\) be an absolutely convergent series in \(\mathbb{K}\) and let \(\sum_{k=1}^\infty a_{\tau(k)}\) be
a reordering. Then \(\sum_{k=1}^\infty
a_{\tau(k)}\) is also absolutely convergent. Moreover, \[\sum_{k=1}^\infty a_{\tau(k)}=\sum_{k=1}^\infty
a_{k}.\]*

*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 \(\sum_{k=1}^\infty |a_{\tau(k)}|\). \(\Box\)

Fabian Gabel and Marcus Waurick. *Well-defined & Wonderful podcast*, marcus-waurick.de.

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