Difference between revisions of "009C Sample Midterm 1, Problem 4"
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!Foundations: | !Foundations: | ||
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| − | | | + | |'''Direct Comparison Test''' |
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| − | |||
|- | |- | ||
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| − | |||
|} | |} | ||
| + | |||
'''Solution:''' | '''Solution:''' | ||
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!Step 1: | !Step 1: | ||
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| − | | | + | |First, we note that |
| + | |- | ||
| + | | <math>\frac{1}{n^23^n}>0</math> | ||
| + | |- | ||
| + | |for all <math>n\ge 1.</math> | ||
| + | |- | ||
| + | |This means that we can use a comparison test on this series. | ||
|- | |- | ||
| − | | | + | |Let <math>a_n=\frac{1}{n^23^n}.</math> |
|} | |} | ||
| Line 31: | Line 36: | ||
!Step 2: | !Step 2: | ||
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| − | | | + | |Let <math>b_n=\frac{1}{n^2}.</math> |
| + | |- | ||
| + | |We want to compare the series in this problem with | ||
| + | |- | ||
| + | | <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{n^2}.</math> | ||
| + | |- | ||
| + | |This is a <math>p</math>-series with <math>p=2.</math> | ||
| + | |- | ||
| + | |Hence, <math>\sum_{n=1}^\infty b_n</math> converges. | ||
| + | |} | ||
| + | |||
| + | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| + | !Step 3: | ||
| + | |- | ||
| + | |Also, we have <math>a_n<b_n</math> since | ||
| + | |- | ||
| + | | <math>\frac{1}{n^23^n}<\frac{1}{n^2}</math> | ||
| + | |- | ||
| + | |for all <math>n\ge 1.</math> | ||
| + | |- | ||
| + | |Therefore, the series <math>\sum_{n=1}^\infty a_n</math> converges | ||
|- | |- | ||
| − | | | + | |by the Direct Comparison Test. |
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| Line 40: | Line 65: | ||
!Final Answer: | !Final Answer: | ||
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| − | | | + | | converges |
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| | | | ||
|} | |} | ||
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 16:22, 12 February 2017
Determine the convergence or divergence of the following series.
Be sure to justify your answers!
| Foundations: |
|---|
| Direct Comparison Test |
Solution:
| Step 1: |
|---|
| First, we note that |
| for all |
| This means that we can use a comparison test on this series. |
| Let |
| Step 2: |
|---|
| Let |
| We want to compare the series in this problem with |
| This is a -series with |
| Hence, converges. |
| Step 3: |
|---|
| Also, we have since |
| for all |
| Therefore, the series converges |
| by the Direct Comparison Test. |
| Final Answer: |
|---|
| converges |