Difference between revisions of "009C Sample Midterm 1, Problem 5"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|  
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|Ratio Test
 
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|-
 
|
 
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'''(a)'''
 
'''(a)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
−
!Step 1:    
+
!Step 1:  
 
|-
 
|-
−
|
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|We first use the Ratio Test to determine the radius of convergence.
 +
|-
 +
|We have
 
|-
 
|-
−
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|&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|}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
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&&\\
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& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
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&&\\
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& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
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&&\\
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& = & \displaystyle{|x|\sqrt{1}}\\
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&&\\
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&=& \displaystyle{|x|.}
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\end{array}</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
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|The Ratio Test tells us this series is absolutely convergent if <math>|x|<1.</math>
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|-
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|Hence, the Radius of Convergence of this series is <math>R=1.</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
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|-
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|Now, we need to determine the interval of convergence.
 +
|-
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|First, note that <math>|x|<1</math> corresponds to the interval <math>(-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>
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|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 4: &nbsp;
 +
|-
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|First, let <math>x=1.</math>
 +
|-
 +
|Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math>
 +
|-
 +
|We note that
 +
|-
 +
|&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.
 +
|-
 +
|Hence, we do not include <math>x=1</math> in the interval.
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|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 5: &nbsp;
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|-
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|Now, let <math>x=-1.</math>
 +
|-
 +
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
 +
|-
 +
|Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
 +
|-
 +
|we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
 +
|-
 +
|Therefore, the series diverges by the <math>n</math>th term test.
 +
|-
 +
|Hence, we do not include <math>x=-1 </math> in the interval.
 +
|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 6: &nbsp;
 
|-
 
|-
−
|
+
|The interval of convergence is <math>(-1,1).</math>
 
|}
 
|}
  

Revision as of 08:17, 13 February 2017

Find the radius of convergence and interval of convergence of the series.

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 \sqrt{n}x^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 (-1)^n \frac{(x-3)^n}{2n+1}}
Foundations:  
Ratio Test

Solution:

(a)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
We have
        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 \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{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{1}}\\ &&\\ &=& \displaystyle{|x|.} \end{array}}
Step 2:  
The Ratio Test tells us this series is absolutely convergent 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 |x|<1.}
Hence, the Radius of Convergence of this series is 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 R=1.}
Step 3:  
Now, we need to determine the interval of convergence.
First, note that 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 |x|<1} corresponds to the interval 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 (-1,1).}
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when 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 R=1.}
Step 4:  
First, let 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 x=1.}
Then, the series becomes 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 \sqrt{n}.}
We note that
        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 \lim_{n\rightarrow \infty} \sqrt{n}=\infty.}
Therefore, the series diverges by the 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 n} th term test.
Hence, we do not include 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 x=1} in the interval.
Step 5:  
Now, let 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 x=-1.}
Then, the series becomes 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 (-1)^n \sqrt{n}.}
Since 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 \lim_{n\rightarrow \infty} \sqrt{n}=\infty,}
we have
        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 \lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=} DNE.
Therefore, the series diverges by the 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 n} th term test.
Hence, we do not include 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 x=-1 } in the interval.
Step 6:  
The interval of convergence is 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 (-1,1).}

(b)

Step 1:  
Step 2:  
Final Answer:  
(a)
(b)

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