009C Sample Midterm 1, Problem 5 - Revision history

Kayla Murray at 05:56, 12 November 2017

2017-11-12T05:56:13Z

Kayla Murray: Replaced content with " Find the radius of convergence and interval of convergence of the series. (a) $\sum_{n=0}^\infty \sqrt{n}x^n$
2017-11-05T01:07:11Z

Replaced content with "<span class="exam"> Find the radius of convergence and interval of convergence of the series. <span class="exam">(a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> <sp..."
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Kayla Murray at 17:22, 18 March 2017

2017-03-18T17:22:51Z

Kayla Murray at 01:59, 27 February 2017

2017-02-27T01:59:39Z

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Kayla Murray at 00:18, 19 February 2017

2017-02-19T00:18:07Z

Kayla Murray at 00:16, 16 February 2017

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Kayla Murray at 22:57, 15 February 2017

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Kayla Murray at 22:49, 15 February 2017

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Kayla Murray at 22:40, 15 February 2017

2017-02-15T22:40:31Z

Kayla Murray at 16:59, 15 February 2017

2017-02-15T16:59:17Z

@@ Line 6: / Line 6: @@
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 164: / Line 164: @@
 |-
 |Let &nbsp;<math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
 |-
 |The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
@@ Line 169: / Line 175: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
 |-
-|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
+|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 |-
 |Also,
@@ Line 204: / Line 210: @@
 \end{array}</math>
 |-
-|Since this limit is a finite number greater than zero, we have
+|Since this limit is a finite number greater than zero,
 |-
-|<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math>&nbsp; diverges by the
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math>&nbsp;
 |-
-|Limit Comparison Test. Therefore, we do not include &nbsp;<math style="vertical-align: -1px">x=2</math>
+|diverges by the Limit Comparison Test.
 |-
-|in our interval.
+|Therefore, we do not include &nbsp;<math style="vertical-align: -1px">x=2</math>&nbsp; in our interval.
 |}

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the radius of convergence and interval of convergence of the series.
-::<span class="exam">a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math>
+<span class="exam">(a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math>
-::<span class="exam">b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 7: / Line 7: @@
 !Foundations: &nbsp;
 |-
-|'''1.''' '''Ratio Test'''
+|'''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>
@@ Line 21: / Line 21: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
 |}

@@ Line 135: / Line 135: @@
 & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
 &&\\
-& = & \displaystyle{|x-3|}
+& = & \displaystyle{|x-3|.}
 \end{array}</math>
 |}
@@ Line 142: / Line 142: @@
 !Step 2: &nbsp;
 |-
-|The Ratio Test tells us this series is absolutely convergent if <math>|x-3|<1.</math>
+|The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|x-3|<1.</math>
 |-
-|Hence, the Radius of Convergence of this series is <math>R=1.</math>
+|Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math>
 |}
@@ Line 152: / Line 152: @@
 |Now, we need to determine the interval of convergence.
 |-
-|First, note that <math>|x-3|<1</math> corresponds to the interval <math>(2,4).</math>
+|First, note that <math style="vertical-align: -5px">|x-3|<1</math> corresponds to the interval <math style="vertical-align: -4px">(2,4).</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>R=1.</math>
+|for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -2px">R=1.</math>
 |}
@@ Line 162: / Line 162: @@
 !Step 4: &nbsp;
 |-
-|First, let <math>x=4.</math>
+|First, let <math style="vertical-align: -1px">x=4.</math>
 |-
 |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
@@ Line 168: / Line 168: @@
 |This is an alternating series.
 |-
-|Let <math>b_n=\frac{1}{2n+1}.</math>.
+|Let <math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>.
 |-
 |The sequence <math>\{b_n\}</math> is decreasing since
@@ Line 174: / Line 174: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
 |-
-|for all <math>n\ge 1.</math>
+|for all <math style="vertical-align: -3px">n\ge 1.</math>
 |-
 |Also,
@@ Line 182: / Line 182: @@
 |Therefore, this series converges by the Alternating Series Test
 |-
-|and we include <math>x=4</math> in our interval.
+|and we include <math style="vertical-align: -1px">x=4</math> in our interval.
 |}
@@ Line 188: / Line 188: @@
 !Step 5: &nbsp;
 |-
-|Now, let <math>x=2.</math>
+|Now, let <math style="vertical-align: -1px">x=2.</math>
 |-
 |Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math>
 |-
-|First, we note that <math>\frac{1}{2n+1}>0</math> for all <math>n\ge 0.</math>
+|First, we note that <math>\frac{1}{2n+1}>0</math> for all <math style="vertical-align: -3px">n\ge 0.</math>
 |-
 |Thus, we can use the Limit Comparison Test.
@@ Line 213: / Line 213: @@
 |<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> diverges by the
 |-
-|Limit Comparison Test. Therefore, we do not include <math>x=2</math>
+|Limit Comparison Test. Therefore, we do not include <math style="vertical-align: -1px">x=2</math>
 |-
 |in our interval.
@@ Line 221: / Line 221: @@
 !Step 6: &nbsp;
 |-
-|The interval of convergence is <math>(2,4].</math>
+|The interval of convergence is <math style="vertical-align: -4px">(2,4].</math>
 |}
@@ Line 228: / Line 228: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval of convergence is <math>(-1,1).</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; 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>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>(2,4].</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -2px">R=1</math> and the interval fo convergence is <math style="vertical-align: -4px">(2,4].</math>
 |}
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 41: / Line 41: @@
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n}|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
+\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
 &&\\
 & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
@@ Line 60: / Line 60: @@
 !Step 2: &nbsp;
 |-
-|The Ratio Test tells us this series is absolutely convergent if <math>|x|<1.</math>
+|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>R=1.</math>
+|Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math>
 |}
@@ Line 70: / Line 70: @@
 |Now, we need to determine the interval of convergence.
 |-
-|First, note that <math>|x|<1</math> corresponds to the interval <math>(-1,1).</math>
+|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>R=1.</math>
+|for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -2px">L=1.</math>
 |}
@@ Line 80: / Line 80: @@
 !Step 4: &nbsp;
 |-
-|First, let <math>x=1.</math>
+|First, let <math style="vertical-align: -1px">x=1.</math>
 |-
 |Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math>
@@ Line 88: / Line 88: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
 |-
-|Therefore, the series diverges by the <math>n</math>th term test.
+|Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test.
 |-
-|Hence, we do not include <math>x=1</math> in the interval.
+|Hence, we do not include <math style="vertical-align: -1px">x=1</math> in the interval.
 |}
@@ Line 96: / Line 96: @@
 !Step 5: &nbsp;
 |-
-|Now, let <math>x=-1.</math>
+|Now, let <math style="vertical-align: -1px">x=-1.</math>
 |-
 |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
@@ Line 104: / Line 104: @@
 |we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math>
 |-
-|Therefore, the series diverges by the <math>n</math>th term test.
+|Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test.
 |-
-|Hence, we do not include <math>x=-1 </math> in the interval.
+|Hence, we do not include <math style="vertical-align: -1px">x=-1 </math> in the interval.
 |}
@@ Line 114: / Line 114: @@
 !Step 6: &nbsp;
 |-
-|The interval of convergence is <math>(-1,1).</math>
+|The interval of convergence is <math style="vertical-align: -4px">(-1,1).</math>
 |}

@@ Line 14: / Line 14: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> 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 <math style="vertical-align: -4px">L>1,</math> 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 <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
 |-
 |'''2.''' If a series absolutely converges, then it also converges.