 index: click on a letter A B C D E F G H I J K L M N O P Q R S T U V W X Y Z A to Z index index: subject areas numbers & symbols sets, logic, proofs geometry algebra trigonometry advanced algebra & pre-calculus calculus advanced topics probability & statistics real world applications multimedia entries www.mathwords.com about mathwords website feedback

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 $\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} = 1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \cdots = \frac{1}{{1 - \left( { - \frac{1}{2}} \right)}} = \frac{2}{3}$