|
|
| (One intermediate revision by the same user not shown) |
| Line 8: |
Line 8: |
| | | | |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Midterm 1, Problem 3 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |'''1.''' A series <math>\sum a_n</math> is '''absolutely convergent''' if
| |
| − | |-
| |
| − | | the series <math>\sum |a_n|</math> converges.
| |
| − | |-
| |
| − | |'''2.''' A series <math>\sum a_n</math> is '''conditionally convergent''' if | |
| − | |-
| |
| − | | the series <math>\sum |a_n|</math> diverges and the series <math>\sum a_n</math> converges.
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:''' | + | [[009C Sample Midterm 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we take the absolute value of the terms in the original series.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -14px">a_n=\frac{(-1)^n}{n}.</math>
| |
| − | |-
| |
| − | |Therefore,
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{n}\bigg|}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math> ).
| |
| − | |-
| |
| − | |Thus, it diverges. Hence, the series
| |
| − | |-
| |
| − | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
| |
| − | |-
| |
| − | |is not absolutely convergent.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 3:
| |
| − | |-
| |
| − | |Now, we need to look back at the original series to see
| |
| − | |-
| |
| − | |if it conditionally converges.
| |
| − | |-
| |
| − | |For
| |
| − | |-
| |
| − | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
| |
| − | |-
| |
| − | |we notice that this series is alternating.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
| |
| − | |-
| |
| − | |First, we have
| |
| − | |-
| |
| − | | <math>\frac{1}{n}\ge 0</math>
| |
| − | |-
| |
| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
| |
| − | |-
| |
| − | |The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since
| |
| − | |-
| |
| − | | <math>\frac{1}{n+1}<\frac{1}{n}</math>
| |
| − | |-
| |
| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
| |
| − | |-
| |
| − | |Also,
| |
| − | |-
| |
| − | | <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>
| |
| − | |-
| |
| − | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> converges
| |
| − | |-
| |
| − | |by the Alternating Series Test.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 4:
| |
| − | |-
| |
| − | |Since the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> is not absolutely convergent but convergent,
| |
| − | |-
| |
| − | |this series is conditionally convergent.
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | conditionally convergent (by the p-test and the Alternating Series Test)
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
