009C Sample Final 1, Problem 4 - Revision history

Kayla Murray: Replaced content with " Find the interval of convergence of the following series. :: $\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}$
009C Sample Final 1, P..."

2017-12-03T01:04:13Z

Replaced content with "<span class="exam"> Find the interval of convergence of the following series. ::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> <hr> 009C Sample Final 1, P..."

Kayla Murray at 18:12, 17 March 2017

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Kayla Murray at 21:37, 24 February 2016

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@@ Line 3: / Line 3: @@
 ::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the interval of convergence of the following series.
-::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
+::<math>\sum_{n=1}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 63: / Line 63: @@
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -5px">n^2<(n+1)^2,</math> &nbsp; we have
+|We notice that this series is alternating.
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math>
+|Let &nbsp;<math style="vertical-align: -14px"> b_n=\frac{1}{n^2}.</math>
 |-
-|Thus, &nbsp; <math>\frac{1}{n^2}</math> &nbsp; is decreasing.
+|First, we have
 |-
-|So, &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> &nbsp; converges by the Alternating Series Test.
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{n^2}\ge 0</math>
 |}
@@ Line 81: / Line 95: @@
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
-\displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
+\displaystyle{\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=1}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
 &&\\
-& = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\
+& = & \displaystyle{\sum_{n=1}^{\infty} \frac{1}{n^2}.}\\
 \end{array}</math>
 |-

@@ Line 8: / Line 8: @@
 |'''1. 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> Then,
+|&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; Then,
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">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: -1px">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: -1px">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.''' After you find the radius of convergence, you need to check the endpoints of your interval
 |-
+|
-&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when &nbsp; <math style="vertical-align: -1px">L=1.</math>
 |}
@@ Line 52: / Line 52: @@
 !Step 2: &nbsp;
 |-
-|So, we have <math style="vertical-align: -6px">|x+2|<1.</math> Hence, our interval is <math style="vertical-align: -3px">(-3,-1).</math> But, we still need to check the endpoints of this interval
+|So, we have &nbsp;<math style="vertical-align: -6px">|x+2|<1.</math>&nbsp; Hence, our interval is &nbsp;<math style="vertical-align: -5px">(-3,-1).</math>&nbsp; But, we still need to check the endpoints of this interval
 |-
 |to see if they are included in the interval of convergence.
@@ Line 60: / Line 60: @@
 !Step 3: &nbsp;
 |-
-|First, we let <math style="vertical-align: -1px">x=-1.</math> Then, our series becomes
+|First, we let &nbsp;<math style="vertical-align: -1px">x=-1.</math> &nbsp; Then, our series becomes
 |-
+|
 &nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 |-
-|Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
+|Since &nbsp;<math style="vertical-align: -5px">n^2<(n+1)^2,</math> &nbsp; we have
 |-
-|So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math>
 |}
@@ Line 73: / Line 77: @@
 !Step 4: &nbsp;
 |-
-|Now, we let <math style="vertical-align: -1px">x=-3.</math> Then, our series becomes
+|Now, we let &nbsp; <math style="vertical-align: -1px">x=-3.</math>&nbsp; Then, our series becomes
 |-
+|
@@ Line 88: / Line 92: @@
 !Step 5: &nbsp;
 |-
-|Thus, the interval of convergence for this series is <math>[-3,-1].</math>
+|Thus, the interval of convergence for this series is &nbsp; <math>[-3,-1].</math>
 |}

@@ Line 90: / Line 90: @@
 |Thus, the interval of convergence for this series is <math>[-3,-1].</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the interval of convergence of the following series.
-::::::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
+::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
-|Recall:
+|'''1. 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> Then,
 |-
+|
-:::If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L>1,</math> the series is divergent.
 |-
+|
-:::If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
+&nbsp; &nbsp; &nbsp; &nbsp;If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive.
 |-
 |'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval
 |-
+|
-:::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
 |}
 '''Solution:'''
@@ Line 35: / Line 34: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<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+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\
 &&\\
@@ Line 64: / Line 63: @@
 |-
+|
-::<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 |-
 |Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
@@ Line 77: / Line 76: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
 &&\\
@@ Line 95: / Line 94: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; <math>[-3,-1]</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>[-3,-1]</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 8: / Line 8: @@
 |Recall:
 |-
-|'''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,
+|'''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,
 |-
+|
@@ Line 22: / Line 22: @@
 |-
+|
-::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1</math>.
+::for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math>
 |}
@@ Line 85: / Line 85: @@
 |-
 |Thus, the interval of convergence for this series is <math>[-3,-1].</math>
 |}

@@ Line 44: / Line 44: @@
 & = & \displaystyle{|x+2|(1)^2}\\
 &&\\
-& = & \displaystyle{|x+2|}\\
+& = & \displaystyle{|x+2|.}\\
 \end{array}</math>
 |}
@@ Line 59: / Line 59: @@
 !Step 3: &nbsp;
 |-
-|First, we let <math style="vertical-align: -1px">x=-1</math>. Then, our series becomes <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math>.
+|First, we let <math style="vertical-align: -1px">x=-1</math>. Then, our series becomes <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math>
 |-
-|Since <math style="vertical-align: -5px">n^2<(n+1)^2</math>, we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}</math>. Thus, <math>\frac{1}{n^2}</math> is decreasing.
+|Since <math style="vertical-align: -5px">n^2<(n+1)^2</math>, we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing.
 |-
 |So, <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}</math> converges by the Alternating Series Test.
@@ Line 75: / Line 75: @@
 \displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\
 &&\\
-& = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}}\\
+& = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\
 \end{array}</math>
 |-
@@ Line 84: / Line 84: @@
 !Step 5: &nbsp;
 |-
-|Thus, the interval of convergence for this series is <math>[-3,-1]</math>.
+|Thus, the interval of convergence for this series is <math>[-3,-1].</math>
 |-
+|