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| | <span class="exam">(b) Find the interval of convergence of the above series. | | <span class="exam">(b) Find the interval of convergence of the above series. |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | |'''Ratio Test'''
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| − | |-
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| − | | 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>
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| − | | Then,
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| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We use the Ratio Test to determine the radius of convergence.
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| − | |-
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| − | |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)^{n+1}}{(n+1)}\cdot\frac{n}{(-1)^n(x)^n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n}{n+1}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n}{n+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|.}
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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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| − | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|x|<1.</math>
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| − | |-
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| − | |Hence, the Radius of Convergence of this series is <math style="vertical-align: -1px">R=1.</math>
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| − | |}
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, note that <math style="vertical-align: -5px">|x|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-1,1).</math>
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| − | |-
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| − | |To obtain the interval of convergence, we need to test the endpoints of this interval
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| − | |for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">R=1.</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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| − | |First, let <math style="vertical-align: -1px">x=1.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=1}^\infty (-1)^n \frac{1}{n}.</math>
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| − | |-
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| − | |This is an alternating series.
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| − | |-
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| − | |Let <math style="vertical-align: -15px">b_n=\frac{1}{n}.</math>.
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| − | |-
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| − | |First, we have
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| − | | <math>\frac{1}{n}\ge 0</math>
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| − | |-
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| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
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| − | |The sequence <math>\{b_n\}</math> is decreasing since
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| − | | <math>\frac{1}{n+1}<\frac{1}{n}</math>
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| − | |-
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| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
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| − | |Also,
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| − | | <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n}=0.</math>
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| − | |Therefore, this series converges by the Alternating Series Test
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| − | |-
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| − | |and we include <math style="vertical-align: -1px">x=1</math> in our interval.
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| − | |}
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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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| − | |Now, let <math style="vertical-align: -1px">x=-1.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=1}^\infty \frac{1}{n}.</math>
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| − | |-
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| − | |This is a <math>p</math>-series with <math>p=1.</math> Hence, the series diverges.
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| − | |-
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| − | |Therefore, we do not include <math style="vertical-align: -1px">x=-1</math> in our interval.
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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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| − | |The interval of convergence is <math style="vertical-align: -4px">(-1,1].</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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| − | | '''(a)''' The radius of convergence is <math style="vertical-align: -1px">R=1.</math>
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| − | |-
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| − | | '''(b)''' <math>(-1,1]</math>
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
