|
|
| Line 3: |
Line 3: |
| | ::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> | | ::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |'''1. Ratio Test'''
| |
| − | |-
| |
| − | | Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> Then,
| |
| − | |-
| |
| − | |
| |
| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
| |
| − | |- | |
| − | |
| |
| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
| |
| − | |-
| |
| − | |
| |
| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
| |
| − | |-
| |
| − | |'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval
| |
| − | |-
| |
| − | |
| |
| − | for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:''' | + | [[009C Sample Final 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We proceed using the ratio test to find the interval of convergence. So, we have
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(-1)^{n+1}(x+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{|x+2|(1)^2}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{|x+2|.}\\
| |
| − | \end{array}</math>
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: -5px">(-3,-1).</math> But, we still need to check the endpoints of this interval
| |
| − | |-
| |
| − | |to see if they are included in the interval of convergence.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 3:
| |
| − | |-
| |
| − | |First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes
| |
| − | |-
| |
| − | |
| |
| − | <math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}.</math>
| |
| − | |-
| |
| − | |We notice that this series is alternating.
| |
| − | |-
| |
| − | |Let <math style="vertical-align: -14px"> b_n=\frac{1}{n^2}.</math>
| |
| − | |-
| |
| − | |First, we have
| |
| − | |-
| |
| − | | <math>\frac{1}{n^2}\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)^2}<\frac{1}{n^2}</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^2}=0.</math>
| |
| − | |-
| |
| − | |So, <math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 4:
| |
| − | |-
| |
| − | |Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=1}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2}.}\\
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |This is a convergent series by the p-test.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 5:
| |
| − | |-
| |
| − | |Thus, the interval of convergence for this series is <math>[-3,-1].</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | <math>[-3,-1]</math>
| |
| − | |}
| |
| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
