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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|-
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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)  

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

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