009C Sample Midterm 3, Problem 2 - Revision history

MathAdmin at 16:43, 28 November 2017

2017-11-28T16:43:57Z

MathAdmin at 21:16, 6 February 2016

2016-02-06T21:16:47Z

MathAdmin at 04:03, 27 April 2015

2015-04-27T04:03:12Z

MathAdmin at 04:01, 27 April 2015

2015-04-27T04:01:52Z

MathAdmin at 03:45, 27 April 2015

2015-04-27T03:45:46Z

MathAdmin at 02:56, 27 April 2015

2015-04-27T02:56:45Z

MathAdmin: Created page with "For each the following series find the sum, if it converges. If you think it diverges, explain why. ::(a) (6 points) &..."

2015-04-27T02:54:11Z

Created page with "For each the following series find the sum, if it converges. If you think it diverges, explain why. ::(a) (6 points) &..."

New page

For each the following series find the sum, if it converges. If
you think it diverges, explain why.

::(a) (6 points)      <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) (6 points)      <math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Foundations:  
|-
|One of the important series to know is the '''Geometric series.''' These are series with a common ratio <math style="vertical-align: 0%">r</math> between adjacent terms which are usually written
|-
|
::<math>\sum_{k=0}^{\infty}a_{0}r^{k}.</math>
|-
|These are convergent if <math style="vertical-align: -22%">|r|<1</math>, and divergent if <math style="vertical-align: -22%">|r|\geq1</math>. If it is convergent, we can find the sum by the formula
|-
|
::<math>S=\frac{a_{0}}{1-r},</math> 
|-
|where <math style="vertical-align: -12%">a_{0}</math> is the first term in the series (if the index starts at <math style="vertical-align: 0%">k=2</math> or <math style="vertical-align: 0%">k=6</math>, then "<math style="vertical-align: -12%">a_{0}</math>" is actually the first term <math style="vertical-align: -12%">a_{2}</math> or <math style="vertical-align: -12%">a_{6}</math>, respectively).
|-
|Another common type of series to evaluate is a '''telescoping series''', where the telescoping better describes the partial sums, denoted <math style="vertical-align: -15%">S_k</math>. Most of the time, they are presented as a fraction which requires '''partial fraction decomposition'''.
|-
|This can be accomplished fairly quickly via a shortcut when the factors in the denominator are linear and share the same coefficient on <math style="vertical-align: 0%">n</math>.
|-
|'''Example.''' Suppose we wish to decompose the fraction <math style="vertical-align: -30%">\frac{4}{(n-2)(n+1)}</math>. First, consider the difference
|-
|
::<math>\frac{1}{n-2}-\frac{1}{n-1}</math>
|-
|If we combine this to a common denominator, we find
|-
|
::<math>\frac{1}{n-2}\cdot\frac{n+1}{n+1}-\frac{1}{n+1}\cdot\frac{n-2}{n-2}\,=\,\frac{n+1-(n-2)}{(n-2)(n+1)}\ =\ \frac{3}{(n-2)(n+1)}.</math>
|-
|To have a 1 in the numerator, we would just multiply by <math style="vertical-align: 0%">\frac{1}{3},</math> or the reciprocal of the difference between the two constants. Thus
|-
|
::<math>\begin{array}{rcl}
{\displaystyle \frac{4}{(n-2)(n+1)}} & = & {\displaystyle 4\cdot\frac{1}{(n-2)(n+1)}}\\
\\
& = & 4\cdot{\displaystyle \frac{1}{3}\left(\frac{1}{n-2}-\frac{1}{n+1}\right).}
\end{array}</math>
|-
|Notice the pattern: for any fraction of the form<math>\frac{1}{(x+a)(x+b)}</math> where<math>a<b,</math> we have
|-
|
::<math>{\displaystyle \frac{1}{(x+a)(x+b)}\,=\,\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right).}</math>
|-
|In this manner, we can quickly find that
|-
|
::<math>{\displaystyle \frac{1}{n^{2}-25}\,=\,\frac{1}{(n-5)(n+5)}\,=\,\frac{1}{5-(-5)}\left(\frac{1}{n-5}-\frac{1}{n+5}\right)\,=\,\frac{1}{10}\left(\frac{1}{n-5}-\frac{1}{n+5}\right)}</math>
|-
|As per the so-called telescoping, consider the series defined by
|-
|
::<math>\sum_{n=2}^{\infty}\frac{1}{n^{2}-1}.</math>
|-
|Using the technique above, we can rewrite the series as
|-
|
::<math>\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}\,=\,\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).</math>
|-
|This means that
|-
|
::<math>\begin{array}{cclcl}
S_{2} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)\\
S_{3} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\\
S_{4} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}\right)\\
S_{5} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{5}-\frac{1}{6}\right)\\
\\
\vdots & \vdots & \vdots\\
S_{n} & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right).
\end{array}</math>
|-
|Notice the pattern - each time there are exactly two surviving positive terms, and two surviving negative terms. This is exactly the difference between the two factors in the denominator. If we then take the limit, we find
|-
|
::<math>\sum_{n=2}^{\infty}\frac{1}{n^{2}-1}\,=\,\lim_{n\rightarrow\infty}S_{n}\,=\,\lim_{n\rightarrow\infty}\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right)\,=\,\frac{3}{4}.</math>
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(a):  
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!(b):  
|-
|
|}
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|
|}
[[009C_Sample_Midterm_3|'''Return to Sample Exam''']]

@@ Line 1: / Line 1: @@
-<span class="exam">For each the following series find the sum, if it converges. If
+<span class="exam">For each the following series find the sum, if it converges.
-::<span class="exam">(a) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <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>
+<span class="exam">If you think it diverges, explain why.
 <br>
-::<span class="exam">(b) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}.</math>
+<span class="exam">(b) &nbsp;<math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}</math>
-&nbsp;'''Solution:'''
+[[009C Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 81: / Line 81: @@
 \end{array}</math>
 |-
-|Notice the pattern; each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms. If we then take the limit, we find
+|Again, notice the pattern: each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms in each partial sum. If we then take the limit, we find
 |-
+|
@@ Line 129: / Line 129: @@
 !Final Answer: &nbsp;
 |-
-|The series in (a) converges to <math style="vertical-align: -23%">3/8</math>, while the series in (b) converges to <math style="vertical-align: -21%">3/2</math>.
+|The series in (a) converges to <math style="vertical-align: -21%">3/8</math>, while the series in (b) converges to <math style="vertical-align: -21%">3/2</math>.
 |}
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 63: / Line 63: @@
 |-
+|
-::<math>\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}\,=\,\sum_{n=1}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).</math>
+::<math>\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}\,=\,\sum_{n=2}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).</math>
 |-
 |This means that

@@ Line 45: / Line 45: @@
 \end{array}</math>
 |-
-|Notice the pattern: for any fraction of the form<math>\frac{1}{(x+a)(x+b)}</math> where<math>a<b,</math> we have
+|Notice the pattern: for any fraction of the form <math style="vertical-align: -20%">\frac{1}{(x+a)(x+b)}</math> where <math style="vertical-align: -20%">a<b,</math> we have
 |-
+|
@@ Line 81: / Line 81: @@
 \end{array}</math>
 |-
-|Notice the pattern; each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms. This is ''<u>exactly</u>'' the difference between the two factors <math style="vertical-align: 0%">n-1</math> and <math style="vertical-align: 0%">n+1</math>in the denominator. If we then take the limit, we find
+|Notice the pattern; each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms. If we then take the limit, we find
 |-
+|
@@ Line 100: / Line 100: @@
 |-
+|
-::<math>S\,=\,\frac{a_{0}}{1-r}\,=\,\frac {1/2}{1-(-1/3)}\,=\,\frac{1/2}{4/3}\,=\,\frac{3}{2}.</math>
+::<math>S\,=\,\frac{a_{0}}{1-r}\,=\,\frac {1/2}{1-(-1/3)}\,=\,\frac{1/2}{4/3}\,=\,\frac{3}{8}.</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !(b): &nbsp;
 |-
+|
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
+|The series in (a) converges to <math style="vertical-align: -23%">3/8</math>, while the series in (b) converges to <math style="vertical-align: -21%">3/2</math>.
 |}
 [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

@@ Line 70: / Line 70: @@
 ::<math>\begin{array}{cclcl}
 S_{2} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)\\
 S_{3} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\\
 S_{4} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}\right)\\
 S_{5} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{5}-\frac{1}{6}\right)\\
 \\
 \vdots & \vdots & \vdots\\
-S_{n} & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right).
+S_{k} & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1}\right).
 \end{array}</math>
 |-
-|Notice the pattern - each time there are exactly two surviving positive terms, and two surviving negative terms. This is exactly the difference between the two factors in the denominator. If we then take the limit, we find
+|Notice the pattern; each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms. This is ''<u>exactly</u>'' the difference between the two factors <math style="vertical-align: 0%">n-1</math> and <math style="vertical-align: 0%">n+1</math>in the denominator. If we then take the limit, we find
 |-
+|
@@ Line 88: / Line 91: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !(a): &nbsp;
 |-
+|
 |}

@@ Line 26: / Line 26: @@
 |This can be accomplished fairly quickly via a shortcut when the factors in the denominator are linear and share the same coefficient on <math style="vertical-align: 0%">n</math>.
 |-
-|'''Example.''' Suppose we wish to decompose the fraction <math style="vertical-align: -30%">\frac{4}{(n-2)(n+1)}</math>.  First, consider the difference
+|'''Example.''' Suppose we wish to decompose the fraction <math style="vertical-align: -22%">\frac{4}{(n-2)(n+1)}</math>.  First, consider the difference
 |-
+|
@@ Line 36: / Line 36: @@
 ::<math>\frac{1}{n-2}\cdot\frac{n+1}{n+1}-\frac{1}{n+1}\cdot\frac{n-2}{n-2}\,=\,\frac{n+1-(n-2)}{(n-2)(n+1)}\ =\ \frac{3}{(n-2)(n+1)}.</math>
 |-
-|To have a 1 in the numerator, we would just multiply by <math style="vertical-align: 0%">\frac{1}{3},</math> or the reciprocal of the difference between the two constants. Thus
+|To have a 1 in the numerator, we would just multiply by <math style="vertical-align: -25%">1/3,</math> or the reciprocal of the difference between the two constants. Thus
 |-
+|