Editing Absolute convergence (section)

==Rearrangements and unconditional convergence==
===Real and complex numbers===

When a series of real or complex numbers is absolutely convergent, any rearrangement or reordering of that series' terms will still converge to the same value. This fact is one reason absolutely convergent series are useful: showing a series is absolutely convergent allows terms to be paired or rearranged in convenient ways without changing the sum's value.

The [[Riemann rearrangement theorem]] shows that the converse is also true: every real or complex-valued series whose terms cannot be reordered to give a different value is absolutely convergent. 

===Series with coefficients in more general space===
The term [[unconditional convergence]] is used to refer to a series where any rearrangement of its terms still converges to the same value. For any series with values in a normed abelian group <math>G</math>, as long as <math>G</math> is complete, every series which converges absolutely also converges unconditionally.

Stated more formally:
{{math theorem| Let <math>G</math> be a normed abelian group. Suppose 
<math display="block">\sum_{i=1}^\infty a_i = A \in G, \quad \sum_{i=1}^\infty \|a_i\|<\infty.</math>
If <math>\sigma : \N \to \N</math> is any permutation, then 
<math display="block">\sum_{i=1}^\infty a_{\sigma(i)}=A.</math>}}

For series with more general coefficients, the converse is more complicated. As stated in the previous section, for real-valued and complex-valued series, unconditional convergence always implies absolute convergence. However, in the more general case of a series with values in any normed abelian group <math>G</math>, the converse does not always hold: there can exist series which are not absolutely convergent, yet unconditionally convergent.

For example, in the [[Banach space]] ℓ<sup>∞</sup>, one series which is unconditionally convergent but not absolutely convergent is:
<math display=block>\sum_{n=1}^\infty \tfrac{1}{n} e_n,</math>

where <math>\{e_n\}_{n=1}^{\infty}</math> is an orthonormal basis.  A theorem of [[Aryeh Dvoretzky|A.&nbsp;Dvoretzky]] and [[Claude Ambrose Rogers|C.&nbsp;A.&nbsp;Rogers]] asserts that every infinite-dimensional Banach space has an unconditionally convergent series that is not absolutely convergent.<ref>Dvoretzky, A.; Rogers, C. A. (1950), "Absolute and unconditional convergence in normed linear spaces", Proc. Natl. Acad. Sci. U.S.A. '''36''':192–197.</ref>

===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.]]