Difference between revisions of "009C Sample Midterm 2, Problem 4"

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|First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>  
 
|First, let &nbsp;<math style="vertical-align: -1px">x=0.</math>  
 
|-
 
|-
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty \frac{1}{\sqrt{n}}.</math>
+
|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
 
|We note that this is a &nbsp;<math style="vertical-align: -3px">p</math>-series with &nbsp;<math style="vertical-align: -12px">p=\frac{1}{2}.</math>
 
|We note that this is a &nbsp;<math style="vertical-align: -3px">p</math>-series with &nbsp;<math style="vertical-align: -12px">p=\frac{1}{2}.</math>
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|Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
 
|Now, let &nbsp;<math style="vertical-align: -1px">x=-2.</math>
 
|-
 
|-
|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
+
|Then, the series becomes &nbsp;<math>\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>
 
|-
 
|-
 
|This series is alternating.  
 
|This series is alternating.  
 
|-
 
|-
 
|Let &nbsp;<math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 
|Let &nbsp;<math style="vertical-align: -20px">b_n=\frac{1}{\sqrt{n}}.</math>
 +
|-
 +
|First, we have
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{\sqrt{n}}\ge 0</math>
 +
|-
 +
|for all &nbsp;<math style="vertical-align: -3px">n\ge 1.</math>
 
|-
 
|-
 
|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 
|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since

Revision as of 10:44, 18 March 2017

Find the radius of convergence and interval of convergence of the series.

(a)  

(b)  


Foundations:  
1. Root Test
        Let    be a positive sequence and let  
        Then,
        If    the series is absolutely convergent.

        If    the series is divergent.

        If    the test is inconclusive.

2. Ratio Test
        Let    be a series and  
        Then,

        If    the series is absolutely convergent.

        If    the series is divergent.

        If    the test is inconclusive.


Solution:

(a)

Step 1:  
We begin by applying the Root Test.
We have

       

Step 2:  
This means that as long as    this series diverges.
Hence, the radius of convergence is    and
the interval of convergence is  

(b)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
We have
       
Step 2:  
The Ratio Test tells us this series is absolutely convergent if  
Hence, the Radius of Convergence of this series is  
Step 3:  
Now, we need to determine the interval of convergence.
First, note that    corresponds to the interval  
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  
Step 4:  
First, let  
Then, the series becomes  
We note that this is a  -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=\frac{1}{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 p<1,}   the series diverges.
Hence, we do not include  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 x=0}   in the interval.
Step 5:  
Now, 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 x=-2.}
Then, the series becomes  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}{\sqrt{n}}.}
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}{\sqrt{n}}.}
First, 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 \frac{1}{\sqrt{n}}\ge 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.}
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}{\sqrt{n+1}}<\frac{1}{\sqrt{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}{\sqrt{n}}=0.}
Therefore, the series converges by the Alternating Series Test.
Hence, we include  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 x=-2}   in our interval of convergence.
Step 6:  
The interval of convergence is  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 [-2,0).}


Final Answer:  
    (a)     The radius of convergence is  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 R=0}   and the interval of convergence is  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 \{0\}.}
    (b)     The radius of convergence is  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 R=1}   and the interval of convergence is  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 [-2,0).}

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