009C Sample Final 2, Problem 2 - Revision history

Kayla Murray at 01:22, 3 December 2017

2017-12-03T01:22:48Z

Grad at 16:54, 12 March 2017

2017-03-12T16:54:15Z

Grad at 00:24, 11 March 2017

2017-03-11T00:24:27Z

Grad at 22:54, 10 March 2017

2017-03-10T22:54:14Z

Kayla Murray at 02:06, 5 March 2017

2017-03-05T02:06:03Z

Kayla Murray at 01:53, 5 March 2017

2017-03-05T01:53:39Z

Kayla Murray at 01:49, 5 March 2017

2017-03-05T01:49:22Z

Kayla Murray at 01:46, 5 March 2017

2017-03-05T01:46:25Z

Kayla Murray at 01:13, 5 March 2017

2017-03-05T01:13:41Z

Kayla Murray at 00:46, 5 March 2017

2017-03-05T00:46:59Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 2, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 2, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 3: / Line 3: @@
 <span class="exam">(a) &nbsp;<math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>
-<span class="exam">(b) &nbsp;<math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math>
+<span class="exam">(b) &nbsp;<math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 69: / Line 69: @@
 |If we let &nbsp;<math style="vertical-align: -14px">x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -14px">A=\frac{1}{2}.</math>
 |-
-|If we let &nbsp;<math style="vertical-align: -14px">x=\frac{-1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -14px">B=\frac{-1}{2}.</math>
+|If we let &nbsp;<math style="vertical-align: -14px">x=-\frac{1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -14px">B=-\frac{1}{2}.</math>
 |-
 |So, we have
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{\frac{-1}{2}}{2n+1}}\\
+\displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{-\frac{1}{2}}{2n+1}}\\
 &&\\
 & = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}

@@ Line 34: / Line 34: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>&nbsp; and &nbsp;<math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
 |-
-|So, this is a geometric series with &nbsp;<math style="vertical-align: -14px">r=\frac{-1}{2}.</math>
+|So, this is a geometric series with &nbsp;<math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
 |-
 |Since &nbsp;<math style="vertical-align: -5px">|r|<1,</math>&nbsp; this series converges.

@@ Line 28: / Line 28: @@
 !Step 1: &nbsp;
 |-
-|Let &nbsp;<math>a_n</math>&nbsp; be the &nbsp;<math>n</math>th term of this sum.
+|Let &nbsp;<math style="vertical-align: -3px">a_n</math>&nbsp; be the &nbsp;<math style="vertical-align: 0px">n</math>th term of this sum.
 |-
 |We notice that
 |-
-|&nbsp; &nbsp; &nbsp, &nbsp; <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>&nbsp; and &nbsp;<math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>&nbsp; and &nbsp;<math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
 |-
-|So, this is a geometric series with &nbsp;<math>r=\frac{-1}{2}.</math>
+|So, this is a geometric series with &nbsp;<math style="vertical-align: -14px">r=\frac{-1}{2}.</math>
 |-
-|Since &nbsp;<math>|r|<1,</math>&nbsp; this series converges.
+|Since &nbsp;<math style="vertical-align: -5px">|r|<1,</math>&nbsp; this series converges.
 |}
@@ Line 63: / Line 63: @@
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
 |-
-|If we multiply this equation by &nbsp;<math>(2x-1)(2x+1),</math>&nbsp; we get
+|If we multiply this equation by &nbsp;<math style="vertical-align: -5px">(2x-1)(2x+1),</math>&nbsp; we get
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp;<math>1=A(2x+1)+B(2x-1).</math>
 |-
-|If we let &nbsp;<math>x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math>A=\frac{1}{2}.</math>
+|If we let &nbsp;<math style="vertical-align: -14px">x=\frac{1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -14px">A=\frac{1}{2}.</math>
 |-
-|If we let &nbsp;<math>x=\frac{-1}{2},</math>&nbsp; we get &nbsp;<math>B=\frac{-1}{2}.</math>
+|If we let &nbsp;<math style="vertical-align: -14px">x=\frac{-1}{2},</math>&nbsp; we get &nbsp;<math style="vertical-align: -14px">B=\frac{-1}{2}.</math>
 |-
 |So, we have
@@ Line 107: / Line 107: @@
 |If we compare &nbsp;<math>s_1,s_2,s_3,</math>&nbsp; we notice a pattern.
 |-
-|We have &nbsp;<math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math>
+|We have
 |}
@@ Line 125: / Line 127: @@
 \end{array}</math>
 |-
-|Since the partial sums converge,  the series converges and the sum of the series is &nbsp;<math>\frac{1}{2}.</math>
+|Since the partial sums converge,  the series converges and the sum of the series is &nbsp;<math style="vertical-align: -15px">\frac{1}{2}.</math>
 |}

@@ Line 59: / Line 59: @@
 !Step 1: &nbsp;
 |-
+|We begin by using partial fraction decomposition. Let
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
 |-
+|If we multiply this equation by &nbsp;<math>(2x-1)(2x+1),</math>&nbsp; we get
 |}
@@ Line 69: / Line 83: @@
 !Step 2: &nbsp;
 |-
+|Now, we look at the partial sums, &nbsp;<math>s_n</math>&nbsp; of this series.
 |-
+|Since the partial sums converge,  the series converges and the sum of the series is &nbsp;<math>\frac{1}{2}.</math>
 |}
@@ Line 80: / Line 134: @@
 |&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;<math>\frac{8}{3}</math>
 |-
-|&nbsp;&nbsp; '''(b)'''
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp;<math>\frac{1}{2}</math>
 |}
 [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
+|'''1.''' The sum of a convergent geometric series is &nbsp; <math>\frac{a}{1-r}</math>
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; where &nbsp;<math style="vertical-align: 0px">r</math>&nbsp; is the ratio of the geometric series
 |-
+|&nbsp; &nbsp; &nbsp; &nbsp; and &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; is the first term of the series.
 |-
+|'''2.''' The &nbsp;<math style="vertical-align: 0px">n</math>th partial sum, &nbsp;<math style="vertical-align: -3px">s_n</math>&nbsp; for a series &nbsp;<math>\sum_{n=1}^\infty a_n </math>&nbsp; is defined as
 |-
+|
 |}

@@ Line 1: / Line 1: @@
 <span class="exam"> For each of the following series, find the sum if it converges. If it diverges, explain why.
-<span class="exam">(a) &nbsp;<math>4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>
+<span class="exam">(a) &nbsp;<math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math>
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math>