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Absolute Convergence
Absolutely Convergent

Describes a series that converges when all terms are replaced by their absolute values. To see if a series converges absolutely, replace any subtraction in the series with addition. If the new series converges, then the original series converges absolutely.

Note: Any series that converges absolutely is itself convergent.

 

Definition:    The series \(\sum\limits_{n = 1}^\infty {{a_n}} \) is absolutely convergent if the series \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) converges.
   
Example:

Determine if  \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots \)  is absolutely convergent.

   
Solution:

To find out, consider the series  \[\sum\limits_{n = 1}^\infty {\left| {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \right|} = \sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \]

 

This is an infinite geometric series with ratio \(r = \frac{1}{2}\), so we know that it converges as shown. \[\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{{1 - \frac{1}{2}}} = 2\]

As a result, we can conclude that \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \) converges absolutely.

   
Note: The series \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots \) and \(\sum\limits_{n = 1}^\infty {\frac{1}{{{2^n}}}} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \) converge to different sums. In fact, \(\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \) is an infinite geometric series with ratio \(r = - \frac{1}{2}\), so \[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots = \frac{1}{{1 - \left( { - \frac{1}{2}} \right)}} = \frac{2}{3}\]

 

See also

Convergence tests

 


  this page updated 19-jul-17
Mathwords: Terms and Formulas from Algebra I to Calculus
written, illustrated, and webmastered by Bruce Simmons
Copyright © 2000 by Bruce Simmons
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