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