009C Sample Midterm 2, Problem 5 - Revision history

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← Older revision		Revision as of 19:42, 12 November 2017
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		+	<hr>
		+	[[009C Sample Midterm 2, Problem 5 Solution\|'''<u>Solution</u>''']]

−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Foundations:  ~~
−	\|-
−	~~\|'''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,~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L<1,</math>  the series is absolutely convergent.~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L>1,</math>  the series is divergent.~~
−	\|-
−	\|
−	~~        If  <math style="vertical-align: -4px">L=1,</math>  the test is inconclusive.~~
−	\|}

		+	[[009C Sample Midterm 2, Problem 5 Detailed Solution\|'''<u>Detailed Solution</u>''']]

−	~~'''Solution:'''~~

−	~~'''(a)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|Assume that the power series  <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>  converges.~~
−	\|-
−	~~\|Let  <math style="vertical-align: 0px">R</math>  be the radius of convergence of this power series.~~
−	\|-
−	~~\|We can use the Ratio Test to find  <math style="vertical-align: 0px">R.</math> ~~
−	\|-
−	~~\|Using the Ratio Test, we have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{n\rightarrow \infty} \bigg\| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg\|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\|\frac{c_{n+1}x}{c_n}\bigg\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\|x\|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Since the radius of convergence of the series  <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>  is  <math style="vertical-align: -5px">R,</math>  we have~~
−	\|-
−	~~\|        <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.</math>~~
−	\|-
−	~~\|Using the Ratio Test, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{n\rightarrow \infty} \bigg\| \frac{c_{n+1}2^nx^{n+1}}{c_n2^{n+1}x^n}\bigg\|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\|\frac{c_{n+1}x}{2c_n}\bigg\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\frac{\|x\|}{2}\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Hence, the radius of convergence of this power series is~~
−	\|-
−	~~\|        <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=2R.</math>~~
−	\|-
−	~~\|Therefore, this power series converges.~~
−	\|}
−
−	~~'''(b)'''~~
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 1:  ~~
−	\|-
−	~~\|Assume that the power series  <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>  converges.~~
−	\|-
−	~~\|Let  <math style="vertical-align: 0px">R</math>  be the radius of convergence of this power series.~~
−	\|-
−	~~\|We can use the Ratio Test to find  <math style="vertical-align: 0px">R.</math> ~~
−	\|-
−	~~\|Using the Ratio Test, we have~~
−	\|-
−	\|
−	~~        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{n\rightarrow \infty} \bigg\| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg\|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\|\frac{c_{n+1}x}{c_n}\bigg\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\|x\|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Since the radius of convergence of the series  <math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>  is  <math style="vertical-align: -5px">R,</math>  we have~~
−	\|-
−	~~\|        <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>~~
−	\|}
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Step 2:  ~~
−	\|-
−	~~\|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n(-x)^n .</math>~~
−	\|-
−	~~\|Using the Ratio Test, we have~~
−	\|-
−	~~\|        <math>\begin{array}{rcl}~~
−	~~\displaystyle{\lim_{n\rightarrow \infty} \bigg\| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg\|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg\|\frac{c_{n+1}(-x)}{c_n}\bigg\|}\\~~
−	~~&&\\~~
−	~~& = & \displaystyle{\|x\|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}~~
−	~~\end{array}</math>~~
−	\|-
−	~~\|Hence, the radius of convergence of this power series is~~
−	\|-
−	~~\|        <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=R.</math>~~
−	\|-
−	~~\|Therefore, this power series converges.~~
−	\|}
−
−
−	~~{\| class="mw-collapsible mw-collapsed" style = "text-align:left;"~~
−	~~!Final Answer:  ~~
−	\|-
−	~~\|    '''(a)'''     converges~~
−	\|-
−	~~\|    '''(b)'''     converges~~
−	\|}
	[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]		[[009C_Sample_Midterm_2\|'''<u>Return to Sample Exam</u>''']]

@@ Line 45: / Line 45: @@
 \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
 &&\\
-& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 \end{array}</math>
 |-
 |Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>
 |}
@@ Line 63: / Line 63: @@
 \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}2^nx^{n+1}}{c_n2^{n+1}x^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{2c_n}\bigg|}\\
 &&\\
-& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 \end{array}</math>
 |-
 |Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=2R.</math>
 |-
 |Therefore, this power series converges.
@@ Line 89: / Line 89: @@
 \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}x^{n+1}}{c_nx^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}x}{c_n}\bigg|}\\
 &&\\
-& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 \end{array}</math>
 |-
 |Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>
 |}
@@ Line 107: / Line 107: @@
 \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{c_{n+1}(-x)^{n+1}}{c_n(-x)^n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{c_{n+1}(-x)}{c_n}\bigg|}\\
 &&\\
-& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\frac{c_{n+1}}{c_n}.}
+& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 \end{array}</math>
 |-
 |Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=R.</math>
 |-
 |Therefore, this power series converges.

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|If a power series converges, then it has a nonempty interval of convergence.
+|'''Ratio Test'''
 |}
@@ Line 24: / Line 37: @@
 |Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 |-
-|So, the power series
+|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
 |}
@@ Line 34: / Line 56: @@
 !Step 2: &nbsp;
 |-
-|Let &nbsp;<math style="vertical-align: -5px">a\in (-2R,2R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math>&nbsp;
+|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n.</math>
 |-
-|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|&nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math>&nbsp; converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
-|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R),</math>&nbsp;
+|Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=2R.</math>
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R).</math>&nbsp;
+|Therefore, this power series converges.
 |}
@@ Line 55: / Line 81: @@
 |Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 |-
-|So, the power series
+|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp;
+|Since the radius of convergence of the series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; is &nbsp;<math style="vertical-align: -5px">R,</math>&nbsp; we have
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}.</math>
 |}
@@ Line 65: / Line 100: @@
 !Step 2: &nbsp;
 |-
-|Let &nbsp;<math style="vertical-align: -5px">a\in (-R,R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">-a \in (-R,R).</math>&nbsp;
+|Now, we use the Ratio Test to find the radius of convergence of the series <math style="vertical-align: -19px">\sum_{n=0}^\infty c_n(-x)^n .</math>
 |-
-|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Using the Ratio Test, we have
 |-
-|&nbsp;<math>\sum_{n=0}^\infty c_n(-a)^n</math>&nbsp; converges.
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 |-
-|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
+|Hence, the radius of convergence of this power series is
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n(-x)^n</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \frac{c_{n+1}}{c_n}}}=R.</math>
 |-
-|converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R).</math>&nbsp;
+|Therefore, this power series converges.
 |}

@@ Line 36: / Line 36: @@
 |Let &nbsp;<math style="vertical-align: -5px">a\in (-2R,2R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math>&nbsp;
 |-
-|So,
+|Since &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges in the interval &nbsp;<math style="vertical-align: -5px">(-R,R),</math>&nbsp;
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math>
+|&nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math>&nbsp; converges.
 |-
-|converges by assumption.
+|Since &nbsp;<math style="vertical-align: 0px">a</math>&nbsp; was an arbitrary number in the interval &nbsp;<math style="vertical-align: -5px">(-2R,2R),</math>&nbsp;
 |-
 |&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n</math>
@@ Line 53: / Line 51: @@
 !Step 1: &nbsp;
 |-
+|Assume that the power series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges.
 |}
@@ Line 59: / Line 65: @@
 !Step 2: &nbsp;
 |-
+|Let &nbsp;<math style="vertical-align: -5px">a\in (-R,R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">-a \in (-R,R).</math>&nbsp;
 |}

@@ Line 34: / Line 34: @@
 !Step 2: &nbsp;
 |-
+|Let &nbsp;<math style="vertical-align: -5px">a\in (-2R,2R).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math>&nbsp;
 |}

@@ Line 46: / Line 46: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges
 |-
-|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; converges
 |}
 [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

@@ Line 10: / Line 10: @@
 !Foundations: &nbsp;
 |-
-|A geometric series &nbsp;<math>\sum_{n=0}^{\infty} ar^n</math>&nbsp; converges if &nbsp;<math style="vertical-align: -6px">|r|<1.</math>
 |}
@@ Line 20: / Line 20: @@
 !Step 1: &nbsp;
 |-
-|First, we notice that &nbsp;<math>\sum_{n=0}^\infty c_nx^n</math>&nbsp; is a geometric series.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}
@@ Line 56: / Line 33: @@
 !Step 1: &nbsp;
 |-
-|First, we notice that &nbsp;<math>\sum_{n=0}^\infty c_nx^n</math>&nbsp; is a geometric series.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 2: &nbsp;
 |-
+|
 |}