Difference between revisions of "009C Sample Midterm 1, Problem 3"

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!Step 1:    
 
!Step 1:    
 
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|-
−
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|First, we take the absolute value of the terms in the original series.
 
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|-
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|Let <math>a_n=\frac{(-1)^n}{n}.</math>
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|-
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|Therefore,
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{n}\bigg|}\\
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&&\\
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& = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.}
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\end{array}</math>
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|This series is the harmonic series (or <math>p</math>-series with <math>p=1</math>).
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|-
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|So, it diverges. Hence the series
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math>
 
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|is not absolutely convergent.
 
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Revision as of 16:05, 12 February 2017

Determine whether the following series converges absolutely, conditionally or whether it diverges.

Be sure to justify your answers!

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}{n}}


Foundations:  
1. A 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 a_n} is absolutely convergent if
        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 |a_n|} converges.
2. A 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 a_n} is conditionally convergent if
        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 |a_n|} diverges and
        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 a_n} converges.

Solution:

Step 1:  
First, we take the absolute value of the terms in the original 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 a_n=\frac{(-1)^n}{n}.}
Therefore,
        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{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{n}\bigg|}\\ &&\\ & = & \displaystyle{\sum_{n=1}^\infty \frac{1}{n}.} \end{array}}
Step 2:  
This series is the harmonic series (or 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 ).
So, it diverges. Hence 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}{n}}
is not absolutely convergent.



Final Answer:  

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