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| Line 10: |
Line 10: |
| | !Foundations: | | !Foundations: |
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| − | |A geometric series <math>\sum_{n=0}^{\infty} ar^n</math> converges if <math style="vertical-align: -6px">|r|<1.</math> | + | | |
| | |} | | |} |
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| Line 20: |
Line 20: |
| | !Step 1: | | !Step 1: |
| | |- | | |- |
| − | |First, we notice that <math>\sum_{n=0}^\infty c_nx^n</math> is a geometric series. | + | | |
| − | |-
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| − | |We have <math style="vertical-align: -1px">r=x.</math>
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| − | |-
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| − | |Since this series converges,
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| − | |-
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| − | | <math>|r|=|x|<1.</math>
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| | |} | | |} |
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| | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" |
| | !Step 2: | | !Step 2: |
| − | |-
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| − | |The series <math>\sum_{n=0} c_n\bigg(\frac{x}{2}\bigg)^n</math> is also a geometric series.
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| − | |-
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| − | |For this series, <math style="vertical-align: -13px">r=\frac{x}{2}.</math>
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| − | |-
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| − | |Now, we notice
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| | |- | | |- |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{|r|} & = & \displaystyle{\bigg|\frac{x}{2}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{|x|}{2}}\\
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| − | &&\\
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| − | & < & \displaystyle{\frac{1}{2}}
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| − | \end{array}</math>
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| − | |-
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| − | |since <math style="vertical-align: -5px">|x|<1.</math>
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| − | |-
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| − | | Since <math style="vertical-align: -5px">|r|<1,</math> this series converges.
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| | |} | | |} |
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| | !Step 1: | | !Step 1: |
| | |- | | |- |
| − | |First, we notice that <math>\sum_{n=0}^\infty c_nx^n</math> is a geometric series. | + | | |
| − | |-
| |
| − | |We have <math style="vertical-align: -1px">r=x.</math>
| |
| − | |-
| |
| − | |Since this series converges,
| |
| − | |-
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| − | | <math>|r|=|x|<1.</math>
| |
| | |} | | |} |
| | | | |
| | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" |
| | !Step 2: | | !Step 2: |
| − | |-
| |
| − | |The series <math>\sum_{n=0}^\infty c_n(-x)^n</math> is also a geometric series.
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| − | |-
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| − | |For this series, <math style="vertical-align: -1px">r=-x.</math>
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| − | |-
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| − | |Now, we notice
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| | |- | | |- |
| | | | | | |
| − | <math>\begin{array}{rcl}
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| − | \displaystyle{|r|} & = & \displaystyle{|-x|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|}\\
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| − | &&\\
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| − | & < & \displaystyle{1}
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| − | \end{array}</math>
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| − | |-
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| − | |since <math style="vertical-align: -5px">|x|<1.</math>
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| − | |-
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| − | |Since <math style="vertical-align: -5px">|r|<1,</math> this series converges.
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| | |} | | |} |
| | | | |
Revision as of 15:45, 23 April 2017
If 
converges, does it follow that the following series converges?
(a) 
(b) 
Solution:
(a)
(b)
| ExpandFinal Answer:
|
| (a) converges (by the geometric series test)
|
| (b) converges (by the geometric series test)
|
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