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

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'''Solution:'''
 
'''Solution:'''
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'''(b)'''
 
'''(b)'''
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
−
!Step 1:    
+
!Step 1:  
 +
|-
 +
|We first use the Ratio Test to determine the radius of convergence.
 +
|-
 +
|We have
 
|-
 
|-
 
|
 
|
 +
&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-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
 +
&&\\
 +
& = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
 +
&&\\
 +
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
 +
&&\\
 +
& = & \displaystyle{|x-3|}
 +
\end{array}</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 2: &nbsp;
 +
|-
 +
|The Ratio Test tells us this series is absolutely convergent if <math>|x-3|<1.</math>
 +
|-
 +
|Hence, the Radius of Convergence of this series is <math>R=1.</math>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 3: &nbsp;
 +
|-
 +
|Now, we need to determine the interval of convergence.
 +
|-
 +
|First, note that <math>|x-3|<1</math> corresponds to the interval <math>(2,4).</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>
 +
|}
 +
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 4: &nbsp;
 
|-
 
|-
−
|
+
|First, let <math>x=4.</math> 
 +
|-
 +
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math>
 +
|-
 +
|This is an alternating series.
 +
|-
 +
|Let <math>b_n=\frac{1}{2n+1}.</math>.
 +
|-
 +
|The sequence <math>\{b_n\}</math> is decreasing since
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math>
 +
|-
 +
|for all <math>n\ge 1.</math>
 +
|-
 +
|Also,
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
 
|-
 
|-
−
|
+
|Therefore, this series converges by the Alternating Series Test
 
|-
 
|-
−
|
+
|and we include <math>x=4</math> in our interval.
 
|}
 
|}
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
−
!Step 2: &nbsp;
+
!Step 5: &nbsp;
 +
|-
 +
|Now, let <math>x=2.</math>
 +
|-
 +
|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math>
 +
|-
 +
|First, we note that <math>\frac{1}{2n+1}>0</math> for all <math>n\ge 0.</math>
 +
|-
 +
|Thus, we can use the Limit Comparison Test.
 +
|-
 +
|We compare this series with the series <math>\sum_{n=1}^\infty \frac{1}{n},</math>
 +
|-
 +
|which is the harmonic series and divergent.
 
|-
 
|-
−
|  
+
|Now, we have
 
|-
 
|-
 
|
 
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
 +
&&\\
 +
& = & \displaystyle{\frac{1}{2}.}
 +
\end{array}</math>
 +
|-
 +
|Since this limit is a finite number greater than zero, we have
 +
|-
 +
|<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> diverges by the
 
|-
 
|-
−
|
+
|Limit Comparison Test. Therefore, we do not include <math>x=2</math>
 +
|-
 +
|in our interval.
 +
|}
 +
 
 +
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 +
!Step 6: &nbsp;
 
|-
 
|-
−
|
+
|The interval of convergence is <math>(2,4].</math>
 
|}
 
|}
 +
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|'''(a)'''  
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval of convergence is <math>(-1,1).</math>
 
|-
 
|-
−
|'''(b)'''  
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>(2,4].</math>
 
|}
 
|}
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 08:24, 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
Then, the series becomes
We note that
       
Therefore, the series diverges by the 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:  
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}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\ &&\\ & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\ &&\\ & = & \displaystyle{|x-3|} \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-3|<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-3|<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 (2,4).}
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=4.}
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 \frac{1}{2n+1}.}
This is an alternating series.
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 b_n=\frac{1}{2n+1}.} .
The sequence 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 \{b_n\}} is decreasing 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 \frac{1}{2(n+1)+1}<\frac{1}{2n+1}}
for all 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\ge 1.}
Also,
        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} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.}
Therefore, this series converges by the Alternating Series Test
and we 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=4} in our 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=2.}
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 \frac{1}{2n+1}.}
First, 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 \frac{1}{2n+1}>0} for all 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\ge 0.}
Thus, we can use the Limit Comparison Test.
We compare this series with the series 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=1}^\infty \frac{1}{n},}
which is the harmonic series and divergent.
Now, 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{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\ &&\\ & = & \displaystyle{\frac{1}{2}.} \end{array}}

Since this limit is a finite number greater than zero, 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 \sum_{n=0}^\infty \frac{1}{2n+1}} diverges by the
Limit Comparison Test. Therefore, 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=2}
in our 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 (2,4].}


Final Answer:  
    (a)     The radius 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 R=1} and 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)     The radius 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 R=1} and the interval fo 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 (2,4].}

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