Difference between revisions of "009C Sample Midterm 1, Problem 4"
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Kayla Murray (talk | contribs) |
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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> |
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |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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!Final Answer: | !Final Answer: | ||
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− | | | + | | converges |
|- | |- | ||
| | | | ||
|} | |} | ||
[[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 |