Editing Absolute convergence (section)

===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>