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| | <span class="exam">(b) <math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math> | | <span class="exam">(b) <math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math> |
| | + | <hr> |
| | + | (insert picture of handwritten solution) |
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| | + | [[009C Sample Midterm 2, Problem 4 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |'''1.''' '''Root Test'''
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| − | |-
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| − | | Let <math>\{a_n\}</math> be a positive sequence and let <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
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| − | |-
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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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| − | |'''2.''' '''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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| − | |-
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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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| − | |-
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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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| − | |-
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| − | |
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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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| − |
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| − |
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| − | '''Solution:'''
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| − |
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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 begin by applying the Root Test.
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| − | |-
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| − | |We have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|n^nx^n|}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} |n^nx^n|^{\frac{1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} |nx|}\\
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| − | &&\\
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| − | & = & \displaystyle{n|x|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
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| − | &&\\
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| − | & = & \displaystyle{\infty.}
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| − | \end{array}</math>
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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 2:
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| − | |-
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| − | |This means that as long as <math style="vertical-align: -6px">x\ne 0,</math> this series diverges.
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| − | |-
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| − | |Hence, the radius of convergence is <math style="vertical-align: -1px">R=0</math> and
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| − | |-
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| − | |the interval of convergence is <math style="vertical-align: -5px">\{0\}.</math>
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| − | |-
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| − | |
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| − | |}
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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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| − | |We first use the Ratio Test to determine the radius of convergence.
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| − | |-
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| − | |We have
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| − | |-
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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{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x+1|\sqrt{1}}\\
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| − | &&\\
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| − | &=& \displaystyle{|x+1|.}
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| − | \end{array}</math>
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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 2:
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| − | |-
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| − | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -4px">|x+1|<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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| − |
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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, we need to determine the interval of convergence.
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| − | |-
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| − | |First, note that <math style="vertical-align: -4px">|x+1|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-2,0).</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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| − | |-
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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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| − |
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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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| − | |First, let <math style="vertical-align: -1px">x=0.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.</math>
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| − | |-
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| − | |We note that this is a <math style="vertical-align: -3px">p</math>-series with <math style="vertical-align: -12px">p=\frac{1}{2}.</math>
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| − | |-
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| − | |Since <math>p<1,</math> the series diverges.
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| − | |-
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| − | |Hence, we do not include <math style="vertical-align: -1px">x=0</math> in the 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 5:
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| − | |-
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| − | |Now, let <math style="vertical-align: -1px">x=-2.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
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| − | |-
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| − | |This series is alternating.
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| − | |-
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| − | |Let <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
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| − | |-
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| − | |First, we have
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| − | |-
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| − | | <math>\frac{1}{\sqrt{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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| − | |-
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| − | |The sequence <math>\{b_n\}</math> is decreasing since
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| − | |-
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| − | | <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{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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| − | |-
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| − | |Also,
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| − | | <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}=0.</math>
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| − | |Therefore, the series converges by the Alternating Series Test.
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| − | |-
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| − | |Hence, we include <math style="vertical-align: -1px">x=-2</math> in our interval of convergence.
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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 6:
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| − | |-
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| − | |The interval of convergence is <math style="vertical-align: -4px">[-2,0).</math>
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| − | |}
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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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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' The radius of convergence is <math style="vertical-align: -1px">R=0</math> and the interval of convergence is <math>\{0\}.</math>
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| − | |-
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| − | | '''(b)''' The radius of convergence is <math style="vertical-align: -1px">R=1</math> and the interval of convergence is <math style="vertical-align: -4px">[-2,0).</math>
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| − | |}
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| | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
