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| | <span class="exam">(b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math> | | <span class="exam">(b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math> |
| | + | <hr> |
| | + | [[009C Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |'''1.''' '''L'Hôpital's Rule''' | |
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| − | Suppose that <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} f(x)</math> and <math style="vertical-align: -11px">\lim_{x\rightarrow \infty} g(x)</math> are both zero or both <math style="vertical-align: -1px">\pm \infty .</math>
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| − | |-
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| − | If <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}</math> is finite or <math style="vertical-align: -4px">\pm \infty ,</math>
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| − | |-
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| − | then <math style="vertical-align: -19px">\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}\,=\,\lim_{x\rightarrow \infty} \frac{f'(x)}{g'(x)}.</math>
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| − | |-
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| − | |'''2.''' The sum of a convergent geometric series is <math>\frac{a}{1-r}</math>
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| − | |-
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| − | | where <math style="vertical-align: 0px">r</math> is the ratio of the geometric series
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| − | |-
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| − | | and <math style="vertical-align: 0px">a</math> is the first term of the series.
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| − | |}
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| − | '''Solution:'''
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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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| − | |Let
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{y} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n.}
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| − | \end{array}</math>
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| − | |-
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| − | |We then take the natural log of both sides to get
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| − | |-
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| − | | <math>\ln y = \ln\bigg(\lim_{n\rightarrow \infty} \bigg(1-\frac{4}{n}\bigg)^n\bigg).</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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| − | |We can interchange limits and continuous functions.
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| − | |-
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| − | |Therefore, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\ln y} & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(1-\frac{4}{n}\bigg)^n}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} n\ln\bigg(1-\frac{4}{n}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
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| − | |-
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| − | |Hence, we can use L'Hopital's Rule to calculate this limit.
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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 have
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| − | |-
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| − | |
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{n}\bigg)}{\frac{1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(1-\frac{4}{x}\bigg)}{\frac{1}{x}}}\\
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| − | &&\\
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| − | & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(1-\frac{4}{x}\big)}\frac{4}{x^2}}{\big(-\frac{1}{x^2}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-4x}{x-4}}\\
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| − | &&\\
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| − | & = & \displaystyle{-4.}
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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 4:
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| − | |-
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| − | |Since <math>\ln y= -4,</math> we know
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| − | |-
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| − | | <math>y=e^{-4}.</math>
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| − | |-
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| − | |Now, we have
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}} & = & \displaystyle{\frac{\displaystyle{\lim_{n\rightarrow \infty} 1}}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{n-4}{n}\bigg)^n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{e^{-4}}}\\
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| − | &&\\
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| − | & = & \displaystyle{e^4.}
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| − | \end{array}</math>
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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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| − | |First, we not that this is a geometric series with <math style="vertical-align: -14px">r=\frac{1}{4}.</math>
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| − | |-
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| − | |Since <math style="vertical-align: -14px">|r|=\frac{1}{4}<1,</math>
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| − | |-
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| − | |this series converges.
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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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| − | |Now, we need to find the sum of this series.
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| − | |-
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| − | |The first term of the series is <math style="vertical-align: -13px">a_1=\frac{1}{2}.</math>
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| − | |-
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| − | |Hence, the sum of the series is
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{3}.}
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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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| − | !Final Answer:
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| − | |-
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| − | | '''(a)''' <math style="vertical-align: -1px">e^{4}</math>
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| − | |-
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| − | | '''(b)''' <math style="vertical-align: -20px">\frac{2}{3}</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>''']] |
