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| | <span class="exam">(b) <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math> | | <span class="exam">(b) <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math> |
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| | + | (insert picture of handwritten solution) |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
| − | !Foundations:
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| − | |-
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| − | |'''1.''' '''Alternating Series Test'''
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| − | |-
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| − | | Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
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| − | |-
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| − | | Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
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| − | |-
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| − | | converge.
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| − | |-
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| − | |'''2.''' '''Ratio Test''' | |
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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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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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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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| − | |'''3.''' If a series absolutely converges, then it also converges.
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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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| − | |First, we have
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| − | |-
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| − | | <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</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 notice that the series is alternating.
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| − | |-
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| − | |Let <math> 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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| − | |-
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| − | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges by the Alternating Series Test.
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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 begin by using the Ratio 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}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(-2)^{n+1} (n+1)!}{(n+1)^{n+1}} \frac{n^n}{(-2)^n n!}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\
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| − | &&\\
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| − | & = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
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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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| − | |Now, we need to calculate <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
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| − | |-
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| − | |Let <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
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| − | |-
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| − | |Then, taking the natural log of both sides, we get
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}
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| − | \end{array}</math>
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| − | |since we can interchange limits and continuous functions.
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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(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
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| − | &&\\
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| − | & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\bigg(\frac{x}{x+1}\bigg)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{-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 4:
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| − | |-
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| − | |Since <math>\ln y=-1,</math> we know
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| − | | <math>y=e^{-1}.</math>
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| − | |Now, we have
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| − | | <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math>
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| − | |Since <math style="vertical-align: -13px">\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test.
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| − | |Therefore, the series converges.
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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)''' converges (by the Alternating Series Test)
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| − | | '''(b)''' converges (by the Ratio Test)
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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>''']] |
