Difference between revisions of "009C Sample Midterm 2, Problem 3"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
−
|Alternating Series Test
+
|'''1.''' '''Alternating Series Test'''
 
|-
 
|-
−
|Ratio Test
+
|&nbsp; &nbsp; &nbsp; &nbsp; Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; converge.
 +
|-
 +
|'''2.''' '''Ratio Test'''
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; 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>
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Then,
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
 +
|-
 +
|
 +
&nbsp; &nbsp; &nbsp; &nbsp; If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
 +
|-
 +
|'''3.''' If a series absolutely converges, then it also converges.
 
|-
 
|-
 
|
 
|
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 
|-
 
|-
−
|for all <math>n\ge 1.</math>
+
|for all <math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|Also,  
 
|Also,  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we need to calculate <math>\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
+
|Now, we need to calculate <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
 
|-
 
|-
−
|Let <math>y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
+
|Let <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math>
 
|-
 
|-
 
|Then, taking the natural log of both sides, we get
 
|Then, taking the natural log of both sides, we get
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|since we can interchange limits and continuous functions.
 
|since we can interchange limits and continuous functions.
 
|-
 
|-
−
|Now, this limit has the form <math>\frac{0}{0}.</math>
+
|Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math>
 
|-
 
|-
 
|Hence, we can use L'Hopital's Rule to calculate this limit.
 
|Hence, we can use L'Hopital's Rule to calculate this limit.
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|Since <math>\ln y=-1,</math> <math>y=e^{-1}.</math>
+
|Since <math>\ln y=-1,</math> we know
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=e^{-1}.</math>
 
|-
 
|-
 
|Now, we have
 
|Now, we have
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math>
 
|-
 
|-
−
|Since <math>\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test.
+
|Since <math style="vertical-align: -13px">\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test.
 
|-
 
|-
 
|Therefore, the series converges.
 
|Therefore, the series converges.

Revision as of 16:15, 15 February 2017

Determine convergence or divergence:

a)
b)


Foundations:  
1. Alternating Series Test
        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 \{a_n\}} be a positive, decreasing sequence where 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} a_n=0.}
        Then, 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 (-1)^na_n} and 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 (-1)^{n+1}a_n}
        converge.
2. Ratio Test
        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 \sum a_n} be a series and 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.}
        Then,

        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 L<1,} the 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 L>1,} the series is divergent.

        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 L=1,} the test is inconclusive.

3. If a series absolutely converges, then it also converges.

Solution:

(a)

Step 1:  
First, 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=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.}
Step 2:  
We notice that the 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 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}{\sqrt{n}}} converges by the Alternating Series Test.

(b)

Step 1:  
We begin by using the Ratio Test.
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{(-2)^{n+1} (n+1)!}{(n+1)^{n+1}} \frac{n^n}{(-2)^n n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\ &&\\ & = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.} \end{array}}

Step 2:  
Now, we need to calculate 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} \bigg(\frac{n}{n+1}\bigg)^n.}
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 y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.}
Then, taking the natural log of both sides, we get

        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{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}} \end{array}}

since we can interchange limits and continuous functions.
Now, this limit has the form 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{0}{0}.}
Hence, we can use L'Hopital's Rule to calculate this limit.
Step 3:  
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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\bigg(\frac{x}{x+1}\bigg)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\ &&\\ & = & \displaystyle{-1.} \end{array}}

Step 4:  
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 \ln y=-1,} we know
        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 y=e^{-1}.}
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 \lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.}
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{2}{e}<1,} the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.
Final Answer:  
    (a)     converges
    (b)     converges

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