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

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|First, we take the absolute value of the terms in the original series.  
 
|First, we take the absolute value of the terms in the original series.  
 
|-
 
|-
−
|Let <math>a_n=\frac{(-1)^n}{n}.</math>
+
|Let <math style="vertical-align: -14px">a_n=\frac{(-1)^n}{n}.</math>
 
|-
 
|-
 
|Therefore,
 
|Therefore,
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|This series is the harmonic series (or <math>p</math>-series with <math>p=1</math>).
+
|This series is the harmonic series (or <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -5px">p=1</math>).
 
|-
 
|-
 
|So, it diverges. Hence, the series  
 
|So, it diverges. Hence, the series  
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|Now, we need to look back at the original series to see
 
|Now, we need to look back at the original series to see
 
|-
 
|-
−
|if it is conditionally converges.  
+
|if it conditionally converges.  
 
|-
 
|-
 
|For <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
 
|For <math>\sum_{n=1}^\infty \frac{(-1)^n}{n},</math>
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|we notice that this series is alternating.  
 
|we notice that this series is alternating.  
 
|-
 
|-
−
|Let <math> b_n=\frac{1}{n}.</math>
+
|Let <math style="vertical-align: -14px"> b_n=\frac{1}{n}.</math>
 
|-
 
|-
−
|The sequence <math>\{b_n\}</math> is decreasing since
+
|The sequence <math style="vertical-align: -4px">\{b_n\}</math> is decreasing since
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}<\frac{1}{n}</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}<\frac{1}{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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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>  
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n}=0.</math>  
 
|-
 
|-
−
|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> converges by the Alternating Series Test.
+
|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; converges by the Alternating Series Test.
 
|}
 
|}
  
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!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|Since the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> is not absolutely convergent but convergent,  
+
|Since the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{n}</math> &nbsp; is not absolutely convergent but convergent,  
 
|-
 
|-
 
|this series is conditionally convergent.
 
|this series is conditionally convergent.

Revision as of 09:21, 14 February 2017

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

Be sure to justify your answers!


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 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} ).
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.
Step 3:  
Now, we need to look back at the original series to see
if it conditionally converges.
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 \frac{(-1)^n}{n},}
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}.}
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+1}<\frac{1}{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}{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}{n}}   converges by the Alternating Series Test.
Step 4:  
Since 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 but convergent,
this series is conditionally convergent.


Final Answer:  
        Conditionally convergent

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