View source for 009C Sample Final 1, Problem 3

<span class="exam">Determine whether the following series converges or diverges.

::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math>

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!Foundations: &nbsp; 
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|Recall:
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::'''1. Ratio Test''' 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> Then,
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:::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent. 
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:::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
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:::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
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::'''2.''' If a series absolutely converges, then it also converges. 
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'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1: &nbsp; 
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|We proceed using the ratio test. 
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|We have
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::<math>\begin{array}{rcl}
\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|}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n!}{n!}\frac{n^n}{(n+1)^{n+1}}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n^n}{(n+1)(n+1)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\bigg(\frac{n}{n+1}\bigg)^n\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^n.}\\
\end{array}</math>
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!Step 2: &nbsp;
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|Now, we continue to calculate the limit from Step 1. We have
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::<math>\begin{array}{rcl}
\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}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}e^{\ln(\frac{n}{n+1})^n}}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty}e^{n\ln(\frac{n}{n+1})}}\\
&&\\
& = & \displaystyle{e^{\lim_{n \rightarrow \infty}n\ln(\frac{n}{n+1})}.}\\
\end{array}</math>
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!Step 3: &nbsp;
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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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|First, we write the limit as 
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::<math style="vertical-align: -16px">\lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.</math>
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|Now, we use L'Hopital's Rule to get 
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::<math>\begin{array}{rcl}
\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}}}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n(n+1)}(-n^2)}\\
&&\\
& = & \displaystyle{\lim_{n \rightarrow \infty} \frac{-n}{n+1}}\\
&&\\
& = & \displaystyle{-1.}\\
\end{array}</math>
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!Step 4: &nbsp;
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|We go back to Step 2 and use the limit we calculated in Step 3. 
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|So, we have 
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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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|Thus, the series absolutely converges by the Ratio Test.
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|Since the series absolutely converges, the series also converges.
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer: &nbsp; 
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|&nbsp;&nbsp; The series converges.
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[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]