009C Sample Final 1, Problem 3 - Revision history

Kayla Murray: Replaced content with "Determine whether the following series converges or diverges. :: $\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}$
009C Sample Final 1, Pro..."

2017-12-03T01:00:05Z

Replaced content with "<span class="exam">Determine whether the following series converges or diverges. ::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math> <hr> 009C Sample Final 1, Pro..."

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@@ Line 3: / Line 3: @@
 ::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 3 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 107: / Line 107: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; converges
+|&nbsp; &nbsp; &nbsp; &nbsp; converges (by the Ratio Test)
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 |'''1.'''  '''Ratio Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -7px">\sum a_n</math> be a series and &nbsp; <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> &nbsp; Then,
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp; <math style="vertical-align: -7px">\sum a_n</math> &nbsp; be a series and &nbsp; <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> &nbsp; Then,
 |-
+|

@@ Line 8: / Line 8: @@
 |'''1.'''  '''Ratio Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; 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,
+|&nbsp; &nbsp; &nbsp; &nbsp; Let <math style="vertical-align: -7px">\sum a_n</math> be a series and &nbsp; <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> &nbsp; Then,
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp; <math style="vertical-align: -4px">L<1,</math> &nbsp; the series is absolutely convergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp; <math style="vertical-align: -4px">L>1,</math> &nbsp; the series is divergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp;If &nbsp; <math style="vertical-align: -4px">L=1,</math> &nbsp; the test is inconclusive.
 |-
+|

@@ Line 78: / Line 78: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<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}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)}\\

@@ Line 1: / Line 1: @@
 <span class="exam">Determine whether the following series converges or diverges.
-::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math>
+::<math>\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1.'''  '''Ratio Test'''
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; 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,
 |-
+|
-:::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
 |-
+|
-::'''2.''' If a series absolutely converges, then it also converges.
+'''2.''' If a series absolutely converges, then it also converges.
 |}
 '''Solution:'''
@@ Line 34: / Line 34: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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|}\\
 &&\\
@@ Line 53: / Line 53: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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}\\
 &&\\
@@ Line 72: / Line 72: @@
 |-
+|
-::<math style="vertical-align: -16px">\lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -16px">\lim_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.</math>
 |-
 |Now, we use L'Hopital's Rule to get
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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}}}\\
 &&\\
@@ Line 96: / Line 96: @@
 |-
+|
-::<math>\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}<1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}<1.</math>
 |-
 |Thus, the series absolutely converges by the Ratio Test.
@@ Line 102: / Line 102: @@
 |Since the series absolutely converges, the series also converges.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;

@@ Line 11: / Line 11: @@
 |-
+|
-::If <math style="vertical-align: -1px">L<1</math>, the series is absolutely convergent.
+::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
 |-
+|
-::If <math style="vertical-align: -1px">L>1</math>, the series is divergent.
+::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
 |-
+|
-::If <math style="vertical-align: -1px">L=1</math>, the test is inconclusive.
+::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
 |-
 |'''2.''' If a series absolutely converges, then it also converges.

009C Sample Final 1, Problem 3 - Revision history

Kayla Murray: Replaced content with "Determine whether the following series converges or diverges. ::\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n} 009C Sample Final 1, Pro..."

Kayla Murray at 18:06, 17 March 2017

Kayla Murray at 23:05, 26 February 2017

Kayla Murray at 22:59, 26 February 2017

Kayla Murray at 23:57, 25 February 2017

Kayla Murray at 23:56, 25 February 2017

Kayla Murray at 23:56, 25 February 2017

Kayla Murray at 01:29, 19 April 2016

Kayla Murray at 18:30, 1 March 2016

Kayla Murray at 18:50, 29 February 2016

Kayla Murray: Replaced content with "Determine whether the following series converges or diverges. :: $\sum_{n=0}^{\infty} (-1)^n \frac{n!}{n^n}$
009C Sample Final 1, Pro..."