009C Sample Midterm 2, Problem 4 - Revision history

Kayla Murray at 19:38, 12 November 2017

2017-11-12T19:38:15Z

Kayla Murray: Replaced content with " Find the radius of convergence and interval of convergence of the series. (a) $\sum_{n=1}^\infty n^nx^n$
2017-11-05T01:14:04Z

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=1}^\infty n^nx^n</math> <span cl..."
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Kayla Murray at 17:44, 18 March 2017

2017-03-18T17:44:03Z

Kayla Murray at 17:41, 18 March 2017

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Kayla Murray at 02:12, 27 February 2017

2017-02-27T02:12:06Z

Kayla Murray at 00:20, 19 February 2017

2017-02-19T00:20:27Z

Kayla Murray at 00:40, 16 February 2017

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Kayla Murray at 00:39, 16 February 2017

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Kayla Murray at 17:47, 13 February 2017

2017-02-13T17:47:45Z

Kayla Murray at 17:26, 13 February 2017

2017-02-13T17:26:33Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
 <hr>
-(insert picture of handwritten solution)
+[[009C Sample Midterm 2, Problem 4 Solution|'''<u>Solution</u>''']]
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 128: / Line 128: @@
 |First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>
 |-
-|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
+|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.</math>
 |-
 |We note that this is a &nbsp;<math style="vertical-align: -3px">p</math>-series with &nbsp;<math style="vertical-align: -12px">p=\frac{1}{2}.</math>
@@ Line 142: / Line 142: @@
 |Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
 |-
-|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
+|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
 |-
 |This series is alternating.
 |-
 |Let &nbsp;<math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 |-
 |The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the radius of convergence and interval of convergence of the series.
-<span class="exam">(a) &nbsp;<math>\sum_{n=0}^\infty n^nx^n</math>
+<span class="exam">(a) &nbsp;<math>\sum_{n=1}^\infty n^nx^n</math>
-<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
+<span class="exam">(b) &nbsp;<math>\sum_{n=1}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>

@@ 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 n^nx^n</math>
+<span class="exam">(a) &nbsp;<math>\sum_{n=0}^\infty n^nx^n</math>
-<span class="exam">(b) <math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
+<span class="exam">(b) &nbsp;<math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>
@@ Line 11: / Line 11: @@
 |'''1.''' '''Root Test'''
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> be a positive sequence and let <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; be a positive sequence and let &nbsp;<math style="vertical-align: -12px">\lim_{n\rightarrow \infty} |a_n|^{\frac{1}{n}}=L.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Then,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
+|&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 |-
 |'''2.''' '''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>
+|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -7px">\sum a_n</math>&nbsp; be a series and &nbsp;<math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; Then,
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 |}
@@ Line 69: / Line 69: @@
 !Step 2: &nbsp;
 |-
-|This means that as long as <math style="vertical-align: -6px">x\ne 0,</math> this series diverges.
+|This means that as long as &nbsp;<math style="vertical-align: -6px">x\ne 0,</math>&nbsp; this series diverges.
 |-
-|Hence, the radius of convergence is <math style="vertical-align: -1px">R=0</math> and
+|Hence, the radius of convergence is &nbsp;<math style="vertical-align: -1px">R=0</math>&nbsp; and
 |-
-|the interval of convergence is <math style="vertical-align: -5px">\{0\}.</math>
+|the interval of convergence is &nbsp;<math style="vertical-align: -5px">\{0\}.</math>
 |-
+|
@@ Line 106: / Line 106: @@
 !Step 2: &nbsp;
 |-
-|The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -4px">|x+1|<1.</math>
+|The Ratio Test tells us this series is absolutely convergent if &nbsp;<math style="vertical-align: -4px">|x+1|<1.</math>
 |-
-|Hence, the Radius of Convergence of this series is <math style="vertical-align: -1px">R=1.</math>
+|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 |}
@@ Line 116: / Line 116: @@
 |Now, we need to determine the interval of convergence.
 |-
-|First, note that <math style="vertical-align: -4px">|x+1|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-2,0).</math>
+|First, note that &nbsp;<math style="vertical-align: -4px">|x+1|<1</math>&nbsp; corresponds to the interval &nbsp;<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: -1px">R=1.</math>
+|for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">R=1.</math>
 |}
@@ Line 126: / Line 126: @@
 !Step 4: &nbsp;
 |-
-|First, let <math style="vertical-align: -1px">x=0.</math>
+|First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>
 |-
-|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
+|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 |-
-|We note that this is a <math style="vertical-align: -3px">p</math>-series with <math style="vertical-align: -12px">p=\frac{1}{2}.</math>
+|We note that this is a &nbsp;<math style="vertical-align: -3px">p</math>-series with &nbsp;<math style="vertical-align: -12px">p=\frac{1}{2}.</math>
 |-
-|Since <math>p<1,</math> the series diverges.
+|Since &nbsp;<math>p<1,</math>&nbsp; the series diverges.
 |-
-|Hence, we do not include <math style="vertical-align: -1px">x=0</math> in the interval.
+|Hence, we do not include &nbsp;<math style="vertical-align: -1px">x=0</math>&nbsp; in the interval.
 |}
@@ Line 140: / Line 140: @@
 !Step 5: &nbsp;
 |-
-|Now, let <math style="vertical-align: -1px">x=-2.</math>
+|Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
 |-
-|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
+|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
 |-
 |This series is alternating.
 |-
-|Let <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
+|Let &nbsp;<math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 |-
-|The sequence <math>\{b_n\}</math> is decreasing since
+|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 |-
-|for all <math style="vertical-align: -3px">n\ge 1.</math>
+|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 |-
 |Also,
@@ Line 160: / Line 160: @@
 |Therefore, the series converges by the Alternating Series Test.
 |-
-|Hence, we include <math style="vertical-align: -1px">x=-2</math> in our interval of convergence.
+|Hence, we include &nbsp;<math style="vertical-align: -1px">x=-2</math>&nbsp; in our interval of convergence.
 |}
@@ Line 166: / Line 166: @@
 !Step 6: &nbsp;
 |-
-|The interval of convergence is <math style="vertical-align: -4px">[-2,0).</math>
+|The interval of convergence is &nbsp;<math style="vertical-align: -4px">[-2,0).</math>
 |}
@@ Line 173: / Line 173: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -1px">R=0</math> and the interval of convergence is <math>\{0\}.</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=0</math>&nbsp; and the interval of convergence is &nbsp;<math>\{0\}.</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -1px">R=1</math> and the interval fo convergence is <math style="vertical-align: -4px">[-2,0).</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=1</math>&nbsp; and the interval of convergence is &nbsp;<math style="vertical-align: -4px">[-2,0).</math>
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ 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 n^nx^n</math>
+<span class="exam">(a) <math>\sum_{n=0}^\infty n^nx^n</math>
-::<span class="exam">b) <math>\sum_{n=0}^\infty \frac{(x+1)^n}{\sqrt{n}}</math>

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-| Root Test
+|'''1.''' '''Root Test'''
 |-
-| Ratio Test
+|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> be a positive sequence and let <math style="vertical-align: -12px">\lim_{n\rightarrow \infty} (a_n)^{\frac{1}{n}}=L.</math>
 |-
+|
 |}
@@ Line 28: / Line 51: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-\displaystyle{\lim_{n\rightarrow \infty} \sqrt{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{|n^nx^n|}}\\
+\displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|a_n|}} & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt[n]{|n^nx^n|}}\\
 &&\\
 & = & \displaystyle{\lim_{n\rightarrow \infty} |n^nx^n|^{\frac{1}{n}}}\\
@@ Line 38: / Line 61: @@
 & = & \displaystyle{|x|\lim_{n\rightarrow \infty} n}\\
 &&\\
-& = & \displaystyle{\infty}
+& = & \displaystyle{\infty.}
 \end{array}</math>
 |}
@@ Line 45: / Line 68: @@
 !Step 2: &nbsp;
 |-
-|This means that as long as <math>x\ne 0,</math> this series diverges.
+|This means that as long as <math style="vertical-align: -6px">x\ne 0,</math> this series diverges.
 |-
-|Hence, the radius of convergence is <math>R=0</math> and
+|Hence, the radius of convergence is <math style="vertical-align: -1px">R=0</math> and
 |-
-|the interval of convergence is <math>\{0\}.</math>
+|the interval of convergence is <math style="vertical-align: -5px">\{0\}.</math>
 |-
+|
@@ Line 63: / Line 86: @@
 |-
 |&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{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
+\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
 &&\\
 & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
@@ Line 82: / Line 105: @@
 !Step 2: &nbsp;
 |-
-|The Ratio Test tells us this series is absolutely convergent if <math>|x+1|<1.</math>
+|The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -4px">|x+1|<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: -1px">R=1.</math>
 |}
@@ Line 92: / Line 115: @@
 |Now, we need to determine the interval of convergence.
 |-
-|First, note that <math>|x+1|<1</math> corresponds to the interval <math>(-2,0).</math>
+|First, note that <math style="vertical-align: -4px">|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>R=1.</math>
+|for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">R=1.</math>
 |}
@@ Line 102: / Line 125: @@
 !Step 4: &nbsp;
 |-
-|First, let <math>x=0.</math>
+|First, let <math style="vertical-align: -1px">x=0.</math>
 |-
 |Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 |-
-|We note that this is a <math>p</math>-series with <math>p=\frac{1}{2}.</math>
+|We note that this is a <math style="vertical-align: -3px">p</math>-series with <math style="vertical-align: -12px">p=\frac{1}{2}.</math>
 |-
 |Since <math>p<1,</math> the series diverges.
 |-
-|Hence, we do not include <math>x=0</math> in the interval.
+|Hence, we do not include <math style="vertical-align: -1px">x=0</math> in the interval.
 |}
@@ Line 116: / Line 139: @@
 !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 (-1)^n \frac{1}{\sqrt{n}}.</math>
@@ Line 122: / Line 145: @@
 |This series is alternating.
 |-
-|Let <math>b_n=\frac{1}{\sqrt{n}}.</math>
+|Let <math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 |-
 |The sequence <math>\{b_n\}</math> is decreasing since
@@ Line 128: / Line 151: @@
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 |-
-|for all <math>n\ge 1.</math>
+|for all <math style="vertical-align: -3px">n\ge 1.</math>
 |-
 |Also,
@@ Line 136: / Line 159: @@
 |Therefore, the series converges by the Alternating Series Test.
 |-
-|Hence, we include <math>x=-2</math> in our interval of convergence.
+|Hence, we include <math style="vertical-align: -1px">x=-2</math> in our interval of convergence.
 |}
@@ Line 142: / Line 165: @@
 !Step 6: &nbsp;
 |-
-|The interval of convergence is <math>[-2,0).</math>
+|The interval of convergence is <math style="vertical-align: -4px">[-2,0).</math>
 |}
@@ Line 149: / Line 172: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=0</math> and the interval of convergence is <math>\{0\}.</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -1px">R=0</math> and the interval of convergence is <math>\{0\}.</math>
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>[-2,0).</math>
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math style="vertical-align: -1px">R=1</math> and the interval fo convergence is <math style="vertical-align: -4px">[-2,0).</math>
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 63: / Line 63: @@
 |-
 |&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}|\frac{a_{n+1}}{a_n}|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
 &&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
+& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
 &&\\
-& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
+& = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\
 &&\\
-& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
+& = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\
 &&\\
-& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
+& = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\
 &&\\
-& = & \displaystyle{|x|\sqrt{1}}\\
+& = & \displaystyle{|x+1|\sqrt{1}}\\
 &&\\
-&=& \displaystyle{|x|.}
+&=& \displaystyle{|x+1|.}
 \end{array}</math>
 |}
@@ Line 82: / Line 82: @@
 !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>|x+1|<1.</math>
 |-
 |Hence, the Radius of Convergence of this series is <math>R=1.</math>
@@ Line 92: / Line 92: @@
 |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>|x+1|<1</math> corresponds to the interval <math>(-2,0).</math>
 |-
 |To obtain the interval of convergence, we need to test the endpoints of this interval
@@ Line 102: / Line 102: @@
 !Step 4: &nbsp;
 |-
-|First, let <math>x=1.</math>
+|First, let <math>x=0.</math>
 |-
-|Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math>
+|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 |-
-|We note that
+|We note that this is a <math>p</math>-series with <math>p=\frac{1}{2}.</math>
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
+|Since <math>p<1,</math> the series diverges.
 |-
-|Therefore, the series diverges by the <math>n</math>th term test.
+|Hence, we do not include <math>x=0</math> in the interval.
 |}
@@ Line 118: / Line 116: @@
 !Step 5: &nbsp;
 |-
-|Now, let <math>x=-1.</math>
+|Now, let <math>x=-2.</math>
 |-
-|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 |-
-|Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
+|for all <math>n\ge 1.</math>
 |-
-|we have
+|Also,
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}=0.</math>
 |-
-|Therefore, the series diverges by the <math>n</math>th term test.
+|Therefore, the series converges by the Alternating Series Test.
 |-
-|Hence, we do not include <math>x=-1 </math> in the interval.
+|Hence, we include <math>x=-2</math> in our interval of convergence.
 |}
@@ Line 136: / Line 142: @@
 !Step 6: &nbsp;
 |-
-|The interval of convergence is <math>(-1,1).</math>
+|The interval of convergence is <math>[-2,0).</math>
 |}
@@ Line 145: / Line 151: @@
 |&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=0</math> and the interval of convergence is <math>\{0\}.</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>R=1</math> and the interval fo convergence is <math>[-2,0).</math>
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]