|
|
| Line 3: |
Line 3: |
| | ::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math> | | ::<math>\sum_{n=1}^\infty \frac{3^n}{n}</math> |
| | | | |
| | + | <hr> |
| | + | [[009C Sample Midterm 2, Problem 2 Solution|'''<u>Solution</u>''']] |
| | | | |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
| |
| − | |-
| |
| − | |'''Direct Comparison Test''' | |
| − | |-
| |
| − | | Let <math>\{a_n\}</math> and <math>\{b_n\}</math> be positive sequences where <math style="vertical-align: -3px">a_n\le b_n</math>
| |
| − | |-
| |
| − | | for all <math style="vertical-align: -3px">n\ge N</math> for some <math style="vertical-align: -3px">N\ge 1.</math>
| |
| − | |-
| |
| − | | '''1.''' If <math>\sum_{n=1}^\infty b_n</math> converges, then <math>\sum_{n=1}^\infty a_n</math> converges.
| |
| − | |-
| |
| − | | '''2.''' If <math>\sum_{n=1}^\infty a_n</math> diverges, then <math>\sum_{n=1}^\infty b_n</math> diverges.
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we note that
| |
| − | |-
| |
| − | | <math>\frac{3^n}{n}>0</math>
| |
| − | |-
| |
| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
| |
| − | |-
| |
| − | |This means that we can use a comparison test on this series.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -13px">a_n=\frac{3^n}{n}.</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
| |
| − | |-
| |
| − | |We want to compare the series in this problem with
| |
| − | |-
| |
| − | | <math>\sum_{n=1}^\infty \frac{1}{n}.</math>
| |
| − | |-
| |
| − | |This is the harmonic series (or <math style="vertical-align: -4px">p</math>-series with <math style="vertical-align: -4px">p=1.</math> )
| |
| − | |-
| |
| − | |Hence, <math>\sum_{n=1}^\infty b_n</math> diverges.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 3:
| |
| − | |-
| |
| − | |Also, we have <math style="vertical-align: -4px">b_n<a_n</math> since
| |
| − | |-
| |
| − | | <math>\frac{1}{n}<\frac{3^n}{n}</math>
| |
| − | |-
| |
| − | | for all <math style="vertical-align: -3px">n\ge 1.</math>
| |
| − | |-
| |
| − | |Therefore, the series <math>\sum_{n=1}^\infty a_n</math> diverges
| |
| − | |-
| |
| − | |by the Direct Comparison Test.
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | diverges (by the Direct Comparison Test)
| |
| − | |}
| |
| | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
