Difference between revisions of "009C Sample Final 3, Problem 4"

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!Foundations:    
 
!Foundations:    
 
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|'''1.''' '''Ratio Test'''
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|-
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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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|-
 
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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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|-
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|'''2.''' If a series absolutely converges, then it also converges.
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|-
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|'''3.''' '''Alternating Series Test'''
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; be a positive, decreasing sequence where &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; Then, &nbsp;<math>\sum_{n=1}^\infty (-1)^na_n</math>&nbsp; and &nbsp;<math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
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|&nbsp; &nbsp; &nbsp; &nbsp; converge.
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
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|We begin by using the Ratio Test.
 
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|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{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!}\bigg|}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \frac{(2n)!}{n!}\bigg|}\\
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\
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&&\\
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& = & \displaystyle{0.}
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\end{array}</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|Since
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,</math>
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|-
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|the series is absolutely convergent by the Ratio Test.
 
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|Therefore, the series converges.
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
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|For
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math>
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|-
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|we notice that this series is alternating.
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|-
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|Let &nbsp;<math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math>
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|-
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|The sequence &nbsp;<math style="vertical-align: -5px">\{b_n\}</math>&nbsp; is decreasing since
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
 
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Also,
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n+1}=0.</math>
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|-
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|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math> &nbsp; converges
 
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|by the Alternating Series Test.
 
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp;&nbsp; '''(a)'''  
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|&nbsp;&nbsp; '''(a)''' &nbsp;&nbsp; converges
 
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|&nbsp;&nbsp; '''(b)'''  
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|&nbsp;&nbsp; '''(b)''' &nbsp;&nbsp; converges
 
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|}
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 13:28, 5 March 2017

Determine if the following series converges or diverges. Please give your reason(s).

(a)  

(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=1}^{+\infty} (-1)^n\frac{1}{n+1}}

Foundations:  
1. 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.

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


Solution:

(a)

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{(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \frac{(2n)!}{n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\ &&\\ & = & \displaystyle{0.} \end{array}}

Step 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 \lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,}
the series is absolutely convergent by the Ratio Test.
Therefore, the series converges.

(b)

Step 1:  
For
        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\frac{1}{n+1},}
we notice that 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}{n+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}{n+2}<\frac{1}{n+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 0.}
Step 2:  
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}{n+1}=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 (-1)^n\frac{1}{n+1}}   converges
by the Alternating Series Test.


Final Answer:  
   (a)    converges
   (b)    converges

Return to Sample Exam

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