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Absolute convergence
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===Proof of the theorem=== For any <math>\varepsilon > 0,</math> we can choose some <math>\kappa_\varepsilon, \lambda_\varepsilon \in \N,</math> such that: <math display=block>\begin{align} \text{ for all } N > \kappa_\varepsilon &\quad \sum_{n=N}^\infty \|a_n\| < \tfrac{\varepsilon}{2} \\ \text{ for all } N > \lambda_\varepsilon &\quad \left\|\sum_{n=1}^N a_n - A\right\| < \tfrac{\varepsilon}{2} \end{align}</math> Let <math display=block>\begin{align} N_\varepsilon &=\max \left\{\kappa_\varepsilon, \lambda_\varepsilon \right\} \\ M_{\sigma,\varepsilon} &= \max \left\{\sigma^{-1}\left(\left\{ 1, \ldots, N_\varepsilon \right\}\right)\right\} \end{align}</math> where <math>\sigma^{-1}\left(\left\{1, \ldots, N_\varepsilon\right\}\right) = \left\{\sigma^{-1}(1), \ldots, \sigma^{-1}\left(N_\varepsilon\right)\right\}</math> so that <math>M_{\sigma,\varepsilon}</math> is the smallest natural number such that the list <math>a_{\sigma(1)}, \ldots, a_{\sigma\left(M_{\sigma,\varepsilon}\right)}</math> includes all of the terms <math>a_1, \ldots, a_{N_\varepsilon}</math> (and possibly others). Finally for any [[integer]] <math> N > M_{\sigma,\varepsilon}</math> let <math display=block>\begin{align} I_{\sigma,\varepsilon} &= \left\{ 1,\ldots,N \right\}\setminus \sigma^{-1}\left(\left \{ 1, \ldots, N_\varepsilon \right \}\right) \\ S_{\sigma,\varepsilon} &= \min \sigma\left(I_{\sigma,\varepsilon}\right) = \min \left\{\sigma(k) \ : \ k \in I_{\sigma,\varepsilon}\right\} \\ L_{\sigma,\varepsilon} &= \max \sigma\left(I_{\sigma,\varepsilon}\right) = \max \left\{\sigma(k) \ : \ k \in I_{\sigma,\varepsilon}\right\} \\ \end{align}</math> so that <math display="block">\begin{align} \left\|\sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)}\right\| &\leq \sum_{i \in I_{\sigma,\varepsilon}} \left\|a_{\sigma(i)}\right\| \\ &\leq \sum_{j = S_{\sigma,\varepsilon}}^{L_{\sigma,\varepsilon}} \left\|a_j\right\| && \text{ since } \sigma(I_{\sigma,\varepsilon}) \subseteq \left\{S_{\sigma,\varepsilon}, S_{\sigma,\varepsilon} + 1, \ldots, L_{\sigma,\varepsilon}\right\} \\ &\leq \sum_{j = N_\varepsilon + 1}^{\infty} \left\|a_j\right\| && \text{ since } S_{\sigma,\varepsilon} \geq N_{\varepsilon} + 1 \\ &< \frac{\varepsilon}{2} \end{align}</math> and thus <math display=block>\begin{align} \left\|\sum_{i=1}^N a_{\sigma(i)}-A \right\| &= \left\| \sum_{i \in \sigma^{-1}\left(\{ 1,\dots,N_\varepsilon \}\right)} a_{\sigma(i)} - A + \sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\| \\ &\leq \left\|\sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \left\|\sum_{i\in I_{\sigma,\varepsilon}} a_{\sigma(i)} \right\| \\ &< \left\|\sum_{j=1}^{N_\varepsilon} a_j - A \right\| + \frac{\varepsilon}{2}\\ &< \varepsilon \end{align}</math> This shows that <math display=block>\text{ for all } \varepsilon > 0, \text{ there exists } M_{\sigma,\varepsilon}, \text{ for all } N > M_{\sigma,\varepsilon} \quad \left\|\sum_{i=1}^N a_{\sigma(i)} - A\right\| < \varepsilon,</math> that is: <math display=block>\sum_{i=1}^\infty a_{\sigma(i)} = A.</math> [[Q.E.D.]]
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