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

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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{3^n}{n}>0</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{3^n}{n}>0</math>
 
|-
 
|-
|for all <math>n\ge 1.</math>
+
|for all <math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|This means that we can use a comparison test on this series.
 
|This means that we can use a comparison test on this series.
 
|-
 
|-
|Let <math>a_n=\frac{3^n}{n}.</math>
+
|Let <math style="vertical-align: -13px">a_n=\frac{3^n}{n}.</math>
 
|}
 
|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|Let <math>b_n=\frac{1}{n}.</math>
+
|Let <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
 
|-
 
|-
 
|We want to compare the series in this problem with
 
|We want to compare the series in this problem with
Line 44: Line 44:
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{1}{n}.</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty \frac{1}{n}.</math>
 
|-
 
|-
|This is the harmonic series (or <math>p</math>-series with <math>p=1.</math>)
+
|This is the harmonic series (or <math style="vertical-align: -4px">p</math>-series with <math style="vertical-align: -4px">p=1.</math>)
 
|-
 
|-
 
|Hence, <math>\sum_{n=1}^\infty b_n</math> diverges.
 
|Hence, <math>\sum_{n=1}^\infty b_n</math> diverges.
Line 52: Line 52:
 
!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
|-
 
|-
|Also, we have <math>b_n<a_n</math> since
+
|Also, we have <math style="vertical-align: -4px">b_n<a_n</math> since
 
|-
 
|-
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n}<\frac{3^n}{n}</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n}<\frac{3^n}{n}</math>
 
|-
 
|-
| for all <math>n\ge 1.</math>
+
| for all <math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|Therefore, the series <math>\sum_{n=1}^\infty a_n</math> diverges
 
|Therefore, the series <math>\sum_{n=1}^\infty a_n</math> diverges

Revision as of 16:02, 15 February 2017

Determine convergence or divergence:

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{3^n}{n}}


Foundations:  
Direct Comparison 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\}} 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 \{b_n\}} be positive sequences 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 a_n\le b_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 N} for some 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.}
        1. 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 \sum_{n=1}^\infty b_n} converges, 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 a_n} converges.
        2. 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 \sum_{n=1}^\infty a_n} diverges, 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 b_n} diverges.

Solution:

Step 1:  
First, we note that
        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{3^n}{n}>0}
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.}
This means that we can use a comparison test on this 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{3^n}{n}.}
Step 2:  
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}.}
We want to compare the series in this problem 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 \sum_{n=1}^\infty \frac{1}{n}.}
This 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.} )
Hence, 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 b_n} diverges.
Step 3:  
Also, 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 b_n<a_n} 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}<\frac{3^n}{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.}
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 a_n} diverges
by the Direct Comparison Test.


Final Answer:  
        diverges

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