Editing Absolute convergence (section)

==== 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>