Difference between revisions of "009C Sample Midterm 2, Problem 4"

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|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&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|}\\
+
\displaystyle{\lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n}|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
+
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
+
& = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
+
& = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
+
& = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{|x|\sqrt{1}}\\
+
& = & \displaystyle{|x+1|\sqrt{1}}\\
 
&&\\
 
&&\\
−
&=& \displaystyle{|x|.}
+
&=& \displaystyle{|x+1|.}
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|The Ratio Test tells us this series is absolutely convergent if <math>|x|<1.</math>
+
|The Ratio Test tells us this series is absolutely convergent if <math>|x+1|<1.</math>
 
|-
 
|-
 
|Hence, the Radius of Convergence of this series is <math>R=1.</math>
 
|Hence, the Radius of Convergence of this series is <math>R=1.</math>
Line 92: Line 92:
 
|Now, we need to determine the interval of convergence.  
 
|Now, we need to determine the interval of convergence.  
 
|-
 
|-
−
|First, note that <math>|x|<1</math> corresponds to the interval <math>(-1,1).</math>
+
|First, note that <math>|x+1|<1</math> corresponds to the interval <math>(-2,0).</math>
 
|-
 
|-
 
|To obtain the interval of convergence, we need to test the endpoints of this interval
 
|To obtain the interval of convergence, we need to test the endpoints of this interval
Line 102: Line 102:
 
!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|First, let <math>x=1.</math>  
+
|First, let <math>x=0.</math>  
 
|-
 
|-
−
|Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math>
+
|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
−
|We note that
+
|We note that this is a <math>p</math>-series with <math>p=\frac{1}{2}.</math>
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
+
|Since <math>p<1,</math> the series diverges.
 
|-
 
|-
−
|Therefore, the series diverges by the <math>n</math>th term test.
+
|Hence, we do not include <math>x=0</math> in the interval.
−
|-
 
−
|Hence, we do not include <math>x=1</math> in the interval.
 
 
|}
 
|}
  
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!Step 5: &nbsp;
 
!Step 5: &nbsp;
 
|-
 
|-
−
|Now, let <math>x=-1.</math>
+
|Now, let <math>x=-2.</math>
 +
|-
 +
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
 +
|-
 +
|This series is alternating.
 +
|-
 +
|Let <math>b_n=\frac{1}{\sqrt{n}}.</math>
 +
|-
 +
|The sequence <math>\{b_n\}</math> is decreasing since
 
|-
 
|-
−
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 
|-
 
|-
−
|Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
+
|for all <math>n\ge 1.</math>
 
|-
 
|-
−
|we have
+
|Also,
 
|-
 
|-
−
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{\sqrt{n}}=0.</math>
 
|-
 
|-
−
|Therefore, the series diverges by the <math>n</math>th term test.
+
|Therefore, the series converges by the Alternating Series Test.
 
|-
 
|-
−
|Hence, we do not include <math>x=-1 </math> in the interval.
+
|Hence, we include <math>x=-2</math> in our interval of convergence.
 
|}
 
|}
  
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!Step 6: &nbsp;
 
!Step 6: &nbsp;
 
|-
 
|-
−
|The interval of convergence is <math>(-1,1).</math>
+
|The interval of convergence is <math>[-2,0).</math>
 
|}
 
|}
  
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=0</math> and the interval of convergence is <math>\{0\}.</math>
 
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is <math>R=0</math> and the interval of convergence is <math>\{0\}.</math>
 
|-
 
|-
−
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>(2,4].</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>[-2,0).</math>
 
|}
 
|}
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 09:47, 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 n^nx^n}
b)


Foundations:  
Root Test
Ratio Test


Solution:

(a)

Step 1:  
We begin by applying the Root Test.
We have

       

Step 2:  
This means that as long as this series diverges.
Hence, the radius of convergence is and
the interval of convergence is

(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}|\frac{a_{n+1}}{a_n}|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(x+1)^{n+1}}{\sqrt{n+1}}\frac{\sqrt{n}}{(x+1)^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (x+1)\frac{\sqrt{n}}{\sqrt{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} |x+1|\frac{\sqrt{n}}{\sqrt{n+1}}}\\ &&\\ & = & \displaystyle{|x+1|\lim_{n\rightarrow \infty} \sqrt{\frac{n}{n+1}}}\\ &&\\ & = & \displaystyle{|x+1|\sqrt{\lim_{n\rightarrow \infty} \frac{n}{n+1}}}\\ &&\\ & = & \displaystyle{|x+1|\sqrt{1}}\\ &&\\ &=& \displaystyle{|x+1|.} \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|<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|<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 (-2,0).}
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=0.}
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}{\sqrt{n}}.}
We note that this is 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 p} -series with 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 p=\frac{1}{2}.}
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 p<1,} the series diverges.
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=0} 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=-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 (-1)^n \frac{1}{\sqrt{n}}.}
This series is alternating.
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}{\sqrt{n}}.}
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}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}}
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}{\sqrt{n}}=0.}
Therefore, the series converges by the Alternating Series Test.
Hence, 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=-2} in our interval of convergence.
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 [-2,0).}


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=0} 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 \{0\}.}
    (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 [-2,0).}

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