Editing Absolute convergence (section)

==Relation to convergence==

If <math>G</math> is [[Complete (topology)|complete]] with respect to the metric <math>d,</math> then every absolutely convergent series is convergent. The proof is the same as for complex-valued series: use the completeness to derive the Cauchy criterion for convergence—a series is convergent if and only if its tails can be made arbitrarily small in norm—and apply the triangle inequality.<!-- too HOWTOish? -->

In particular, for series with values in any [[Banach space]], absolute convergence implies convergence. The converse is also true: if absolute convergence implies convergence in a normed space, then the space is a Banach space.<!--Banach spaces are specific here.-->

If a series is convergent but not absolutely convergent, it is called [[conditionally convergent]]. An example of a conditionally convergent series is the [[alternating harmonic series]]. Many standard tests for divergence and convergence, most notably including the [[ratio test]] and the [[root test]], demonstrate absolute convergence. This is because a [[power series]] is absolutely convergent on the interior of its disk of convergence.{{Efn|Here, the disk of convergence is used to refer to all points whose distance from the center of the series is less than the radius of convergence. That is, the disk of convergence is made up of all points for which the power series converges.}}

===Proof that any absolutely convergent series of complex numbers is convergent===

Suppose that <math display=inline>\sum \left|a_k\right|, a_k \in \Complex</math> is convergent. Then equivalently, <math display=inline>\sum \left[ \operatorname{Re}\left(a_k\right)^2 + \operatorname{Im}\left(a_k\right)^2 \right]^{1/2}</math> is convergent, which implies that <math display=inline>\sum \left|\operatorname{Re}\left(a_k\right)\right|</math> and <math display=inline>\sum\left|\operatorname{Im}\left(a_k\right)\right|</math> converge by termwise comparison of non-negative terms. It suffices to show that the convergence of these series implies the convergence of <math display=inline>\sum \operatorname{Re}\left(a_k\right)</math> and <math display=inline>\sum \operatorname{Im}\left(a_k\right),</math> for then, the convergence of <math display=inline>\sum a_k=\sum \operatorname{Re}\left(a_k\right) + i \sum \operatorname{Im}\left(a_k\right)</math> would follow, by the definition of the convergence of complex-valued series.

The preceding discussion shows that we need only prove that convergence of <math display=inline>\sum \left|a_k\right|, a_k\in\R</math> implies the convergence of <math display=inline>\sum a_k.</math>

Let <math display=inline>\sum \left|a_k\right|, a_k\in\R</math> be convergent.  Since <math>0 \leq a_k + \left|a_k\right| \leq 2\left|a_k\right|,</math> we have
<math display=block>0 \leq \sum_{k = 1}^n (a_k + \left|a_k\right|) \leq \sum_{k = 1}^n 2\left|a_k\right|.</math>
Since <math display=inline>\sum 2\left|a_k\right|</math> is convergent, <math display=inline>s_n=\sum_{k = 1}^n \left(a_k + \left|a_k\right|\right)</math> is a [[Sequence#Bounded|bounded]] [[Sequence#Increasing and decreasing|monotonic]] [[sequence]] of partial sums, and <math display=inline>\sum \left(a_k + \left|a_k\right|\right)</math> must also converge. Noting that <math display=inline>\sum a_k = \sum \left(a_k + \left|a_k\right|\right) - \sum \left|a_k\right|</math> is the difference of convergent series, we conclude that it too is a convergent series, as desired.

==== Alternative proof using the Cauchy criterion and triangle inequality ====

By applying the Cauchy criterion for the convergence of a complex series, we can also prove this fact as a simple implication of the [[triangle inequality]].<ref>{{Cite book|url=https://archive.org/details/1979RudinW|title=Principles of Mathematical Analysis|last=Rudin|first=Walter|publisher=McGraw-Hill|year=1976|isbn=0-07-054235-X|location=New York|pages=71–72}}</ref>  By the [[Cauchy's convergence test|Cauchy criterion]], <math display=inline>\sum |a_i|</math> converges if and only if for any <math>\varepsilon > 0,</math> there exists <math>N</math> such that <math display=inline>\left|\sum_{i=m}^n \left|a_i\right| \right| = \sum_{i=m}^n |a_i| < \varepsilon</math> for any <math>n > m \geq N.</math> But the triangle inequality implies that <math display=inline>\big|\sum_{i=m}^n a_i\big| \leq \sum_{i=m}^n |a_i|,</math> so that <math display=inline>\left|\sum_{i=m}^n a_i\right| < \varepsilon</math> for any <math>n > m \geq N,</math> which is exactly the Cauchy criterion for <math display=inline>\sum a_i.</math>

===Proof that any absolutely convergent series in a Banach space is convergent===

The above result can be easily generalized to every [[Banach space]] <math>(X, \|\,\cdot\,\|).</math>  Let <math display="inline">\sum x_n</math> be an absolutely convergent series in <math>X.</math> As <math display=inline>\sum_{k=1}^n\|x_k\|</math> is a [[Cauchy sequence]] of real numbers, for any <math>\varepsilon > 0</math> and large enough [[natural number]]s <math>m > n</math> it holds:
<math display=block>\left| \sum_{k=1}^m \|x_k\| - \sum_{k=1}^n \|x_k\| \right| = \sum_{k=n+1}^m \|x_k\| < \varepsilon.</math>

By the triangle inequality for the norm {{math|ǁ⋅ǁ}}, one immediately gets:
<math display=block>\left\|\sum_{k=1}^m x_k - \sum_{k=1}^n x_k\right\| = \left\|\sum_{k=n+1}^m x_k\right\| \leq \sum_{k=n+1}^m \|x_k\| < \varepsilon,</math>
which means that <math display=inline>\sum_{k=1}^n x_k</math> is a Cauchy sequence in <math>X,</math> hence the series is convergent in <math>X.</math><ref>{{citation
|last = Megginson|first = Robert E.|author-link = Robert Megginson
|title = An introduction to Banach space theory
|series = Graduate Texts in Mathematics
|volume = 183 
|publisher = Springer-Verlag
|location = New York 
|year = 1998
|isbn = 0-387-98431-3
|page = 20
}} (Theorem 1.3.9)</ref>