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| | <span class="exam"> Find the interval of convergence of the following series. | | <span class="exam"> Find the interval of convergence of the following series. |
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| − | ::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
| + | ::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | |Ratio Test
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| − | |Check endpoints of interval
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| − | |} | |
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| − | '''Solution:'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Final 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 1:
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| − | |-
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| − | |We proceed using the ratio test to find the interval of convergence. So, we have
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| − | ::<math>\begin{array}{rcl}
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| − | \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|\\
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| − | &&\\
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| − | & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+2|(1)^2}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+2|}\\
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| − | \end{array}</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |So, we have <math>|x+2|<1</math>. Hence, our interval is <math>(-3,-1)</math>. But, we still need to check the endpoints of this interval
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| − | |to see if they are included in the interval of convergence.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |First, we let <math>x=-1</math>. Then, our series becomes <math>sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math>.
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| − | |-
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| − | |Since <math>n^2<(n+1)^2</math>, we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}</math>. Thus, <math>\frac{1}{n^2}</math> is decreasing.
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| − | |-
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| − | |So, <math>sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |-
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| − | |Now, we let <math>x=-3</math>. Then, our series becomes
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
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| − | &&\\
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| − | & = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}}\\
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| − | \end{array}</math>
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| − | |-
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| − | |This is a convergent series by the p-test.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 5:
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| − | |-
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| − | |Thus, the interval of convergence for this series is <math>[-3,-1]</math>.
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |-
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| − | | <math>[-3,-1]</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
