Difference between revisions of "009C Sample Midterm 2, Problem 5"

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[[009C Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Foundations: &nbsp;
 
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|'''Ratio Test'''
 
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|&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>
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Then,
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 
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[[009C Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''Solution:'''
 
  
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'''(a)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|Assume that the power series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges.
 
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|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
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|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 
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|Using the Ratio Test, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\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|}\\
 
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&&\\
 
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& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 
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\end{array}</math>
 
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|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
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|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>
 
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|Using the Ratio Test, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\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|}\\
 
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&&\\
 
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& = & \displaystyle{\frac{|x|}{2}\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 
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\end{array}</math>
 
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|Hence, the radius of convergence of this power series is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{2}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=2R.</math>
 
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|Therefore, this power series converges.
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|Assume that the power series &nbsp;<math style="vertical-align: -19px">\sum_{n=0}^\infty c_nx^n</math>&nbsp; converges.
 
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|Let &nbsp;<math style="vertical-align: 0px">R</math>&nbsp; be the radius of convergence of this power series.
 
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|We can use the Ratio Test to find &nbsp;<math style="vertical-align: 0px">R.</math>&nbsp;
 
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|Using the Ratio Test, we have
 
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\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|}\\
 
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&&\\
 
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& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 
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\end{array}</math>
 
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|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
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>R=\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}.</math>
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|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>
 
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|Using the Ratio Test, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\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|}\\
 
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&&\\
 
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& = & \displaystyle{|x|\lim_{n\rightarrow \infty}\bigg(\frac{c_{n+1}}{c_n}\bigg).}
 
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\end{array}</math>
 
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|Hence, the radius of convergence of this power series is
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\displaystyle{\lim_{n\rightarrow \infty} \bigg(\frac{c_{n+1}}{c_n}\bigg)}}=R.</math>
 
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|Therefore, this power series converges.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; converges
 
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[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 11:42, 12 November 2017

If  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty c_nx^n}   converges, does it follow that the following series converges?

(a)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n}

(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^\infty c_n(-x)^n }


Solution


Detailed Solution


Return to Sample Exam

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