2017-12-01T19:35:54Z

MathAdmin:

A lighthouse is located on a small island 3km away from the nearest point  <math style="vertical-align: 0px">P</math>  on a straight shoreline and its light makes 4 revolutions per minute. How fast is the beam of light moving along the shoreline on a point 1km away from  <math style="vertical-align: 0px">P?</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations:  
|-
|When we see a problem talking about rates, it is usually a '''related rates''' problem.
|-
|Thus, we treat everything as a function of time, or  <math style="vertical-align: -1px">t.</math>
|-
|We can usually find an equation relating one unknown to another, and then use implicit differentiation.
|-
|Since the problem usually gives us one rate, and from the given info we can usually find the values of
variables at our particular moment in time, we can solve the equation
for the remaining rate.
|}

'''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We can begin this physical word problem by drawing a picture.
|-
|[[File:009A_SF2_5GP.png|left|400px]]
|-
|In the picture, we can consider the distance from the point  <math style="vertical-align: 0px">P</math>  to the spot the light hits the shore to be the variable  <math style="vertical-align: 0px">x.</math>
|-
|By drawing a right triangle with the beam as its hypotenuse, we can see that our variable
 <math style="vertical-align: 0px">x</math>  is related to the angle  <math style="vertical-align: 0px">\theta</math>  by the equation
|-
|       <math>{\displaystyle \tan\theta\ =\ \frac{\textrm{side~opp.}}{\textrm{side~adj. }}\ =\ \frac{x}{3}.}</math>
|-
|This gives us a relation between the two variables.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we use implicit differentiation to find
|-
|       <math>{\displaystyle \sec^{2}\theta\cdot\frac{d\theta}{dt}\ =\ \frac{1}{3}\cdot\frac{dx}{dt}.}</math>
|-
|Rearranging, we have
|-
|       <math>{\displaystyle \frac{dx}{dt}\ =\ 3\sec^{2}\theta\cdot\frac{d\theta}{dt}.}</math>
|-
|Again, everything is a function of time.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|We want to know the rate that the beam is moving along the shore when
we are one km away from the point  <math style="vertical-align: 0px">P.</math>
|-
|This tells us that  <math style="vertical-align: -1px">x=1.</math>
|-
|The problem also tells us that the lighthouse beam is revolving at 4 revolutions
per minute.
|-
|However,  <math style="vertical-align: 0px">\theta</math>  is measured in radians, and there are  <math style="vertical-align: 0px">2\pi</math>  radians in a revolution.
|-
|Thus, we know
|-
|       <math>{\displaystyle \frac{d\theta}{dt}\ =\ 4\cdot2\pi\ =\ 8\pi.}</math>
|-
|Finally, we require secant. Since we know  <math style="vertical-align: -3px">x=1,</math>
|-
|we can solve the triangle to get that the length of the hypotenuse is
|-
|       <math>\sqrt{1^{2}+3^{2}}=\sqrt{10}.</math>
|-
|This means that
|-
|       <math>{\displaystyle \sec\theta\ =\ \frac{1}{\cos\theta}\ =\ \frac{\textrm{hyp.}}{\textrm{side~adj.}}\ =\ \frac{\sqrt{10}}{3}.}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|Now, we can plug in all these values to find
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\frac{dx}{dt}} & = & \displaystyle{3\sec^{2}\theta\cdot\frac{d\theta}{dt}}\\
&&\\
& = & \displaystyle{3\left(\frac{\sqrt{10}}{3}\right)^{2}(8\pi)}\\
&&\\
& = & \displaystyle{3\left(\frac{10}{9}\right)(8\pi)}\\
&&\\
& = & \displaystyle{\frac{80\pi}{3} \text{ km/min.}}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>80\pi\text{ km/min}</math>
|}
[[009A_Sample_Final_2|'''Return to Sample Exam''']]

File:9CSM3P5.jpg

2017-11-29T18:04:00Z

MathAdmin:

007B Sample Midterm 3, Problem 3 Detailed Solution

2017-11-28T16:51:17Z

MathAdmin: Created page with " For a fish that starts life with a length of 1cm and has a maximum length of 30cm, the von Bertalanffy growth model predicts that the growth rate is  ..."

For a fish that starts life with a length of 1cm and has a maximum length of 30cm, the von Bertalanffy growth model predicts that the growth rate is  <math style="vertical-align: 0px">29e^{-t}</math>  cm/year where  <math style="vertical-align: 0px">t</math>  is the age of the fish. What is the average length of the fish over its first five years?
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|The average value of a function  <math style="vertical-align: -5px">f(x)</math>  on an interval  <math style="vertical-align: -5px">[a,b]</math>  is given by
|-
|
::<math>f_{\text{avg}}=\frac{1}{b-a} \int_a^b f(x)~dx.</math>
|}

'''Solution:'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Let  <math style="vertical-align: -5px">L(t)</math>  be the length of the fish at age  <math style="vertical-align: 0px">t.</math>
|-
|Given the information in the problem, we know
|-
|       <math style="vertical-align: -4px">L'(t)=29e^{-t}</math>  and  <math style="vertical-align: -5px">L(0)=1.</math>
|-
|First, we find  <math style="vertical-align: -5px">L(t).</math> 
|-
|We have
|-
|
        <math>\begin{array}{rcl}
\displaystyle{L(t)} & = & \displaystyle{\int L'(t)~dt}\\
&&\\
& = & \displaystyle{\int 29e^{-t}~dt}\\
&&\\
& = & \displaystyle{-29e^{-t}+C.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Next, we need to solve for  <math style="vertical-align: -1px">C.</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{1} & = & \displaystyle{L(0)}\\
&&\\
& = & \displaystyle{-29e^{0}+C}\\
&&\\
& = & \displaystyle{-29+C.}
\end{array}</math>
|-
|So, we get  <math style="vertical-align: -1px">C=30.</math>
|-
|Therefore,
|-
|
       <math>L(t)=-29e^{-t}+30.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Finally, we need to find the average value of  <math style="vertical-align: -5px">L(t)</math>  over the interval  <math style="vertical-align: -5px">[0,5].</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{L_{\text{avg}}} & = & \displaystyle{\frac{1}{5-0}\int_0^5 L(t)~dt}\\
&&\\
& = & \displaystyle{\frac{1}{5}\int_0^5 -29e^{-t}+30~dt}\\
&&\\
& = & \displaystyle{\frac{1}{5}(29e^{-t}+30t)\bigg|_0^5}\\
&&\\
& = & \displaystyle{\frac{1}{5}(29e^{-5}+30(5))-\frac{1}{5}(29e^0+0)}\\
&&\\
& = & \displaystyle{\frac{1}{5}(29e^{-5}+121)\text{ cm}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>\frac{1}{5}(29e^{-5}+121) \text{ cm}</math>
|}
[[007B_Sample_Midterm_3|'''Return to Sample Exam''']]

007B Sample Midterm 3, Problem 3 Solution

2017-11-28T16:50:51Z

MathAdmin: Created page with "center '''Return to Sample Exam'''"

[[File:7BSM3P3.jpg|600px|thumb|center]]

[[007B_Sample_Midterm_3|'''Return to Sample Exam''']]

007B Sample Midterm 3, Problem 3

2017-11-28T16:50:18Z

009C Sample Midterm 3, Problem 5 Detailed Solution

2017-11-28T16:48:38Z

MathAdmin: Created page with " Find the radius of convergence and the interval of convergence of the series. (a)  <math>{\displaystyle \sum_{n=0}^{\infty}}\frac..."

Find the radius of convergence and the interval of convergence
of the series.

(a)  <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}</math>

(b)  <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''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.
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|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{(-1)^{n+1}x^{n+1}}{n+1+1} \frac{n+1}{(-1)^nx^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)x\frac{n+1}{n+2}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{n+2}|x|}\\
&&\\
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n+1}{n+2}}\\
&&\\
& = & \displaystyle{|x| (1)}\\
&&\\
&=& \displaystyle{|x|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-1,1).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">L=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=-1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{n+1}.</math>
|-
|We note that  <math style="vertical-align: -15px">\frac{1}{n+1}>0</math>  for all  <math style="vertical-align: -3px">n\ge 0.</math> 
|-
|Hence, we can use the limit comparison test for this series.
|-
|Let  <math style="vertical-align: -15px"> a_n=\frac{1}{n+1}</math>  and  <math style="vertical-align: -13px">b_n=\frac{1}{n}.</math> 
|-
|Notice that  <math style="vertical-align 1px">b_n>0</math>  for all  <math style="vertical-align: -3px">n\ge 1.</math> 
|-
|The series  <math style="vertical-align: -1 px">\sum_{n=1}^\infty \frac{1}{n} </math>  is a  <math style="vertical-align: -3px">p</math>-series with  <math style="vertical-align: -3px">p=1.</math>  So,  <math style="vertical-align: -1px>\sum_{n=1}^\infty b_n</math>  diverges.
|-
|Now,
|-
|
        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}\frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|Therefore, the series  <math>\sum_{n=0}^\infty \frac{1}{n+1}</math>   diverges by the limit comparison test.
|-
|Hence, we do not include  <math style="vertical-align: -1px">x=-1</math>  in the interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=1.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -16px">b_n=\frac{1}{n+1}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{n+1}\ge 0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|The sequence  <math>\{b_n\}</math>  is decreasing since
|-
|        <math>\frac{1}{(n+1)+1}<\frac{1}{n+1}</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 0.</math>
|-
|Also,
|-
|        <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n+1}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=1</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We first use the Ratio Test to determine the radius of convergence.
|-
|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{(x+1)^{n+1}}{(n+1)^2}\frac{n^2}{(x+1)^n}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(x+1)\frac{n^2}{n^2+2n+1}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{n^2}{n^2+2n+1}}\\
&&\\
& = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \frac{n^2}{n^2+2n+1}}\\
&&\\
& = & \displaystyle{|x+1|.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The Ratio Test tells us this series is absolutely convergent if  <math style="vertical-align: -5px">|x+1|<1.</math>
|-
|Hence, the Radius of Convergence of this series is  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we need to determine the interval of convergence.
|-
|First, note that  <math style="vertical-align: -5px">|x+1|<1</math>  corresponds to the interval  <math style="vertical-align: -4px">(-2,0).</math>
|-
|To obtain the interval of convergence, we need to test the endpoints of this interval
|-
|for convergence since the Ratio Test is inconclusive when  <math style="vertical-align: -2px">R=1.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 4:  
|-
|First, let  <math style="vertical-align: -1px">x=-2.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty (-1)^n \frac{1}{n^2}.</math>
|-
|This is an alternating series.
|-
|Let  <math style="vertical-align: -16px">b_n=\frac{1}{n^2}.</math>.
|-
|First, we have
|-
|       <math>\frac{1}{n^2}\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}{(n+1)^2}<\frac{1}{n^2}</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}{n^2}=0.</math>
|-
|Therefore, this series converges by the Alternating Series Test
|-
|and we include  <math style="vertical-align: -1px">x=-2</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 5:  
|-
|Now, let  <math style="vertical-align: -1px">x=0.</math>
|-
|Then, the series becomes  <math>\sum_{n=0}^\infty \frac{1}{n^2}.</math>
|-
|This is a  <math style="vertical-align: -4px">p</math>-series with  <math style="vertical-align: -3px>p=2.</math> 
|-
|Hence, this series converges.
|-
|Therefore, we do include  <math style="vertical-align: -1px">x=0</math>  in our interval.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 6:  
|-
|The interval of convergence is  <math style="vertical-align: -4px">[-2,0].</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">(-1,1].</math>
|-
|    '''(b)'''     The radius of convergence is  <math style="vertical-align: -2px">R=1</math>  and the interval of convergence is  <math style="vertical-align: -4px">[-2,0].</math>
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 5 Solution

2017-11-28T16:48:09Z

MathAdmin: Created page with "center '''Return to Sample Exam'''"

[[File:9CSM3P5.jpg|600px|thumb|center]]

[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 5

2017-11-28T16:47:54Z

MathAdmin:

009C Sample Midterm 3, Problem 4 Detailed Solution

2017-11-28T16:47:23Z

MathAdmin: Created page with "Test the series for convergence or divergence. (a)  <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\sin\frac{\pi}{n}</math> <..."

Test the series for convergence or divergence.

(a)  <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\sin\frac{\pi}{n}</math>

(b)  <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1)^{n}\cos\frac{\pi}{n}</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''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.
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\sin \bigg(\frac{\pi}{n}\bigg)>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 1.</math>
|-
|So, the series
|-
|        <math>\sum_{n=1}^\infty (-1)^n\sin \bigg(\frac{\pi}{n}\bigg)</math>
|-
|is alternating.
|-
|Let  <math style="vertical-align: -17px">b_n=\sin \bigg(\frac{\pi}{n}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|The sequence  <math style="vertical-align: -5px">\{b_n\}</math>  is decreasing since
|-
|        <math>\sin\bigg(\frac{\pi}{n+1}\bigg)<\sin\bigg(\frac{\pi}{n}\bigg)</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 2.</math>
|-
|Also,
|-
|
       <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty}b_n} & = &\displaystyle{\lim_{n\rightarrow \infty}\sin\bigg(\frac{\pi}{n}\bigg)}\\
&&\\
& = & \displaystyle{\sin(0)}\\
&&\\
& = & \displaystyle{0.}
\end{array} </math>
|-
|Therefore,
|-
|       <math>\sum_{n=1}^\infty (-1)^n\sin \frac{\pi}{n}</math>  
|-
|converges by the Alternating Series Test.

|}

'''(b)'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\cos \bigg(\frac{\pi}{n}\bigg)>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 3.</math>
|-
|So, the series
|-
|        <math>\sum_{n=1}^\infty (-1)^n\cos \bigg(\frac{\pi}{n}\bigg)</math>
|-
|is alternating.
|-
|Also, we have
|-
|
       <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \cos \bigg(\frac{\pi}{n}\bigg)} & = &\displaystyle{\cos (0)}\\
&&\\
& = & \displaystyle{1.}
\end{array} </math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since  <math style="vertical-align: -16px">\lim_{n\rightarrow \infty} \cos \bigg(\frac{\pi}{n}\bigg)\neq 0,</math>  we have
|-
|        <math>\lim_{n\rightarrow \infty} (-1)^n \cos \bigg(\frac{\pi}{n}\bigg)=DNE.</math>
|-
|Therefore, the series diverges by the Divergence Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    (a)     converges (by the Alternating Series Test)
|-
|    (b)     diverges (by the Divergence Test)
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 4 Solution

2017-11-28T16:46:53Z

MathAdmin: Created page with "center '''Return to Sample Exam'''"

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[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 4

2017-11-28T16:46:37Z

MathAdmin:

009C Sample Midterm 3, Problem 3 Detailed Solution

2017-11-28T16:46:07Z

MathAdmin: Created page with "Test if each the following series converges or diverges. Give reasons and clearly state if you are using any standard test. <span clas..."

Test if each the following series converges or diverges.

Give reasons and clearly state if you are using any standard test.

(a)  <math>{\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+1)!}</math>
 

(b)  <math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}</math>
<hr>
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''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,
|-
|
        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.
|-
|'''2.''' If a series absolutely converges, then it also converges.
|-
|'''3.''' '''Limit Comparison Test'''
|-
|        Let  <math>\{a_n\}</math>  and  <math>\{b_n\}</math>  be positive sequences.
|-
|        If  <math style="vertical-align: -16px">\lim_{n\rightarrow \infty} \frac{a_n}{b_n}=L,</math>  where  <math style="vertical-align: -1px">L</math>  is a positive real number,
|-
|        then  <math style="vertical-align: -20px">\sum_{n=1}^\infty a_n</math>  and  <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>  either both converge or both diverge.
|}

'''Solution:'''

'''(a)'''
{| 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{(n+1)!}{(3(n+1)+1)!} \frac{(3n+1)!}{n!}\bigg|}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(n+1)n!}{(3n+4)!} \frac{(3n+1)!}{n!}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(n+1)(3n+1)!}{(3n+4)(3n+3)(3n+2)(3n+1)!}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(3n+4)(3n+3)(3n+2)}}\\
&&\\
& = & \displaystyle{0.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Since  <math style="vertical-align: -4px">0<1,</math>  the series is absolutely convergent by the Ratio Test.
|-
|Therefore, the series converges.
|}

(b)
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|First, we note that
|-
|        <math>\frac{\sqrt{n}}{n^2-3}>0</math>
|-
|for all  <math style="vertical-align: -3px">n\ge 2.</math>
|-
|This means that we can use a comparison test on this series.
|-
|Let  <math style="vertical-align: -14px">a_n=\frac{\sqrt{n}}{n^2-3}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Let  <math style="vertical-align: -17px">b_n=\frac{\sqrt{n}}{n^2}=\frac{1}{n^{\frac{3}{2}}}.</math>
|-
|We want to compare the series in this problem with
|-
|        <math>\sum_{n=1}^\infty b_n=\sum_{n=2}^\infty \frac{1}{n^{\frac{3}{2}}}.</math>
|-
|This is a  <math style="vertical-align: -4px">p</math>-series with  <math style="vertical-align: -14px">p=\frac{3}{2}.</math>
|-
|Hence,  <math>\sum_{n=2}^\infty b_n</math>  converges.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\big(\frac{\sqrt{n}}{n^2-3}\big)}{\bigg(\frac{1}{n^{\frac{3}{2}}}\bigg)}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\sqrt{n})n^{\frac{3}{2}}}{n^2-3}}\\
&&\\
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n^2}{n^2-3}}\\
&&\\
& = & \displaystyle{1.}
\end{array}</math>
|-
|Therefore, the series
|-
|        <math>\sum_{n=2}^{\infty} \frac{\sqrt{n}}{n^2-3}</math>
|-
|converges by the Limit Comparison Test.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    (a)     converges (by the Ratio Test)
|-
|    (b)     converges (by the Limit Comparison Test)
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 3 Solution

2017-11-28T16:45:42Z

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[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 3

2017-11-28T16:45:23Z

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009C Sample Midterm 3, Problem 2 Detailed Solution

2017-11-28T16:44:45Z

MathAdmin: Created page with "For each the following series find the sum, if it converges. If you think it diverges, explain why. (a)  <math..."

For each the following series find the sum, if it converges.

If you think it diverges, explain why.

(a)  <math style="vertical-align: -50%">\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3^{2}}-\frac{1}{2\cdot3^{3}}+\frac{1}{2\cdot3^{4}}-\frac{1}{2\cdot3^{5}}+\cdots </math>
 

(b)  <math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}</math>
<hr>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Background Information:  
|-
|'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math>   with   <math>|r|<1,</math>
|-
|
       <math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
|-
|
'''2.''' For a telescoping series, we find the sum by first looking at the partial sum   <math style="vertical-align: -3px">s_k</math>
|-
|
       and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
|}

'''Solution:'''

'''(a)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|Each term grows by a ratio of  <math style="vertical-align: -13px">\frac{1}{3}</math>  and it reverses sign.
|-
|Thus, there is a common ratio  <math style="vertical-align: -13px">r=-\frac{1}{3}.</math>
|-
|Also, the first term is  <math style="vertical-align: -13px">\frac{1}{2}.</math>  So, we can write the series as a geometric series given by
|-
|
::<math>\sum_{n=0}^{\infty}\,\frac{1}{2}\left(-\frac{1}{3}\right)^n.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Then, the series converges to the sum
|-
|
       <math>\begin{array}{rcl}
\displaystyle{S} & = & \displaystyle{\frac{a}{1-r}}\\
&&\\
& = & \displaystyle{\frac {\frac{1}{2}}{1-(-\frac{1}{3})}}\\
&&\\
& = & \displaystyle{\frac{\frac{1}{2}}{\frac{4}{3}}}\\
&&\\
& = & \displaystyle{\frac{3}{8}.}
\end{array}</math>
|}

'''(b)'''
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 1:  
|-
|We begin by using partial fraction decomposition. Let
|-
|       <math>\frac{3}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
|-
|If we multiply this equation by  <math style="vertical-align: -5px">(2x-1)(2x+1),</math>  we get
|-
|       <math>3=A(2x+1)+B(2x-1).</math>
|-
|If we let  <math style="vertical-align: -14px">x=\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">A=\frac{3}{2}.</math>
|-
|If we let  <math style="vertical-align: -14px">x=-\frac{1}{2},</math>  we get  <math style="vertical-align: -14px">B=-\frac{3}{2}.</math>
|-
|So, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\sum_{n=1}^\infty \frac{3}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{3}{2}}{2n-1}+\frac{-\frac{3}{2}}{2n+1}}\\
&&\\
& = & \displaystyle{\frac{3}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
\end{array}</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 2:  
|-
|Now, we look at the partial sums,  <math>s_n</math>  of this series.
|-
|First, we have
|-
|        <math>s_1=\frac{3}{2}\bigg(1-\frac{1}{3}\bigg).</math>
|-
|Also, we have
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_2} & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
&&\\
& = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{5}\bigg)}
\end{array}</math>
|-
|and
|-
|        <math>\begin{array}{rcl}
\displaystyle{s_3} & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
&&\\
& = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{7}\bigg).}
\end{array}</math>
|-
|If we compare  <math>s_1,s_2,s_3,</math>  we notice a pattern.
|-
|We have
|-
|       <math>s_n=\frac{3}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Step 3:  
|-
|Now, to calculate the sum of this series we need to calculate
|-
|       <math>\lim_{n\rightarrow \infty} s_n.</math>
|-
|We have
|-
|        <math>\begin{array}{rcl}
\displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{3}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
&&\\
& = & \displaystyle{\frac{3}{2}.}
\end{array}</math>
|-
|Since the partial sums converge, the series converges and the sum of the series is  <math style="vertical-align: -15px">\frac{3}{2}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|    '''(a)'''     <math>\frac{3}{8}</math>
|-
|    '''(b)'''     <math>\frac{3}{2}</math>
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

009C Sample Midterm 3, Problem 2 Solution

2017-11-28T16:44:12Z

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009C Sample Midterm 3, Problem 2

2017-11-28T16:43:57Z

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009C Sample Midterm 3, Problem 1 Solution

2017-11-28T16:43:24Z

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File:9CSM3P4.jpg

2017-11-28T16:42:51Z

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File:9CSM3P3.jpg

2017-11-28T16:42:40Z

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