009C Sample Midterm 2, Problem 3 - Revision history

MathAdmin: Replaced content with "Determine convergence or divergence: (a) $\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}$ (b) &nbs..."

2017-11-12T19:39:45Z

Replaced content with "Determine convergence or divergence: (a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math> (b) &nbs..."

MathAdmin at 17:33, 14 April 2017

2017-04-14T17:33:58Z

MathAdmin: Created page with "Determine convergence or divergence: (a) $\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}$ (b)
2017-04-09T19:30:41Z

Created page with "Determine convergence or divergence: (a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math> (b) <m..."

New page
Determine convergence or divergence:

(a)  <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math>

(b)  <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|'''1.''' '''Alternating Series Test'''
|-
|        Let  <math>\{a_n\}</math>  be a positive, decreasing sequence where  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
|-
|        Then,  <math>\sum_{n=1}^\infty (-1)^na_n</math>  and  <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
|-
|        converge.
|-
|'''2.''' '''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,
|-
|
        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.
|-
|
        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.
|-
|
        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.
|-
|'''3.''' If a series absolutely converges, then it also converges.
|-
|
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we have
|-
|        <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|We notice that the series is alternating.
|-
|Let  <math> b_n=\frac{1}{\sqrt{n}}.</math>
|-
|First, we have
|-
|       <math>\frac{1}{\sqrt{n}}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
|-
|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>  converges by the Alternating Series Test.
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using the Ratio Test.
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\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|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\
&&\\
& = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we need to calculate  <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
|-
|Let  <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
|-
|Then, taking the natural log of both sides, we get
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}
\end{array}</math>
|-
|since we can interchange limits and continuous functions.
|-
|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>
|-
|Hence, we can use L'Hopital's Rule to calculate this limit.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
&&\\
& \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)}}\\
&&\\
& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\
&&\\
& = & \displaystyle{-1.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Since  <math>\ln y=-1,</math>  we know
|-
|        <math>y=e^{-1}.</math>
|-
|Now, we have
|-
|        <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math>
|-
|Since  <math style="vertical-align: -13px">\frac{2}{e}<1,</math>  the series is absolutely convergent by the Ratio Test.
|-
|Therefore, the series converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     converges (by the Alternating Series Test)
|-
|    '''(b)'''     converges (by the Ratio Test)
|}
[[009C_Sample_Midterm_2|'''Return to Sample Exam''']]

← Older revision		Revision as of 19:39, 12 November 2017
Line 5:		Line 5:
	<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>

		+	<hr>
		+	[[009C Sample Midterm 2, Problem 3 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|'''1.''' '''Alternating Series Test'''~~
−	\|-
−	~~\|        Let  <math>\{a_n\}</math>  be a positive, decreasing sequence where  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>~~
−	\|-
−	~~\|        Then,  <math>\sum_{n=1}^\infty (-1)^na_n</math>  and  <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>~~
−	\|-
−	~~\|        converge.~~
−	\|-
−	~~\|'''2.''' '''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,~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.~~
−	\|-
−	~~\|'''3.''' If a series absolutely converges, then it also converges.~~
−	\|-
−	\|
−	\|}

		+	[[009C Sample Midterm 2, Problem 3 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|First, we have~~
−	\|-
−	~~\|        <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|We notice that the series is alternating.~~
−	\|-
−	~~\|Let  <math> b_n=\frac{1}{\sqrt{n}}.</math>~~
−	\|-
−	~~\|First, we have~~
−	\|-
−	~~\|       <math>\frac{1}{\sqrt{n}}\ge 0</math>~~
−	\|-
−	~~\|for all  <math style="vertical-align: -3px">n\ge 1.</math>~~
−	\|-
−	~~\|The sequence  <math>\{b_n\}</math>  is decreasing since~~
−	\|-
−	~~\|        <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>~~
−	\|-
−	~~\|for all  <math style="vertical-align: -3px">n\ge 1.</math>~~
−	\|-
−	~~\|Also,~~
−	\|-
−	~~\|        <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>~~
−	\|-
−	~~\|Therefore, the series  <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math>  converges by the Alternating Series Test.~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|We begin by using the Ratio Test.~~
−	\|-
−	~~\|We have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\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\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we need to calculate  <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>~~
−	\|-
−	~~\|Let  <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>~~
−	\|-
−	~~\|Then, taking the natural log of both sides, we get~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|since we can interchange limits and continuous functions.~~
−	\|-
−	~~\|Now, this limit has the form  <math style="vertical-align: -13px">\frac{0}{0}.</math>~~
−	\|-
−	~~\|Hence, we can use L'Hopital's Rule to calculate this limit.~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 3:  ~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \big(\frac{n}{n+1}\big)}{\frac{1}{n}}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \big(\frac{x}{x+1}\big)}{\frac{1}{x}}}\\~~
−	~~&&\\~~
−	~~& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(\frac{x}{x+1}\big)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{-1.}~~
−	~~\end{array}</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 4:  ~~
−	\|-
−	~~\|Since  <math>\ln y=-1,</math>  we know~~
−	\|-
−	~~\|        <math>y=e^{-1}.</math>~~
−	\|-
−	~~\|Now, we have~~
−	\|-
−	~~\|        <math>\lim_{n\rightarrow \infty} \bigg\|\frac{a_{n+1}}{a_n}\bigg\|=2e^{-1}=\frac{2}{e}.</math>~~
−	\|-
−	~~\|Since  <math style="vertical-align: -13px">\frac{2}{e}<1,</math>  the series is absolutely convergent by the Ratio Test.~~
−	\|-
−	~~\|Therefore, the series converges.~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     converges (by the Alternating Series Test)~~
−	\|-
−	~~\|    '''(b)'''     converges (by the Ratio Test)~~
−	\|}
	[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 129: / Line 129: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\
+\displaystyle{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \big(\frac{n}{n+1}\big)}{\frac{1}{n}}}\\
 &&\\
-& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\
+& = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \big(\frac{x}{x+1}\big)}{\frac{1}{x}}}\\
 &&\\
-& \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)}}\\
+& \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\big(\frac{x}{x+1}\big)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\
 &&\\
 & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\

009C Sample Midterm 2, Problem 3 - Revision history

MathAdmin: Replaced content with "Determine convergence or divergence: (a) \sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}} (b) &nbs..."

MathAdmin at 17:33, 14 April 2017

MathAdmin: Replaced content with "Determine convergence or divergence: (a) $\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}$ (b) &nbs..."