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| | <span class="exam">Determine whether the following series converges or diverges. | | <span class="exam">Determine whether the following series converges or diverges. |
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| − | ::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math>
| + | ::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 3 Solution|'''<u>Solution</u>''']] |
| − | |-
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| − | |Review Ratio Test
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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 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Step 1:
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| − | |-
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| − | |We proceed using the ratio test.
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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}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{(-1)^n n!}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n!}{n!}\frac{n^n}{(n+1)^{n+1}}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n^n}{(n+1)(n+1)^n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\bigg(\frac{n}{n+1}\bigg)^n\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\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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| − | {| 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 continue to calculate the limit from Step 1. We have
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| − | |-
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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{n}{n+1}\bigg)^n}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}e^{\ln(\frac{n}{n+1})^n}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty}e^{n\ln(\frac{n}{n+1})}}\\
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| − | &&\\
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| − | & = & \displaystyle{e^{\lim_{n \rightarrow \infty}n\ln(\frac{n}{n+1})}}\\
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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 3:
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| − | |-
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| − | |Now, we need to calculate <math>\lim_{n \rightarrow \infty}n\ln\bigg(\frac{n}{n+1}\bigg)</math>.
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| − | |-
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| − | |First, we write the limit as <math>\lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}</math>.
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| − | |-
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| − | |Now, we use L'Hopital's Rule to get
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| − | |
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| − | ::<math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n \rightarrow \infty}n\ln\bigg(\frac{n}{n+1}\bigg)} & \overset{l'H}{=} & \displaystyle{\lim_{n \rightarrow \infty}\frac{\frac{n+1}{n}\frac{(n+1)-n}{(n+1)^2}}{-\frac{1}{n^2}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n(n+1)}(-n^2)}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{-n}{n+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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| − | |We go back to Step 2 and use the limit we calculated in Step 3.
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| − | |-
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| − | |So, we have
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| − | |-
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| − | |<math>\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}<1</math>.
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| − | |-
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| − | |Thus, the series absolutely converges by the Ratio Test.
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| − | |-
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| − | |Since the series absolutely converges, the series also 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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| − | |-
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| − | |The series converges.
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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>''']] |
