Difference between revisions of "009C Sample Midterm 1, Problem 5 Detailed Solution"

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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{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
+
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}\cdot x^{n+1}}{\sqrt{n}\cdot x^n}\bigg|}\\
 
&&\\
 
&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}\cdot x\bigg|}\\
 
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&&\\
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& = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
+
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\sqrt{\frac{n+1}{n}}\cdot|x|\bigg)}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
 
& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
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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{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
+
\displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\cdot \frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
+
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)\cdot (x-3)\cdot \frac{2n+1}{2n+3}\bigg|}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
+
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(|x-3|\cdot \frac{2n+1}{2n+3}\bigg)}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
 
& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
−
\displaystyle{\lim_{n\rightarrow \infty} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
+
\displaystyle{\lim_{n\rightarrow \infty} \frac{\big(\frac{1}{2n+1}\big)}{\big(\frac{1}{n}\big)}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{\frac{1}{2}.}
 
& = & \displaystyle{\frac{1}{2}.}

Latest revision as of 13:26, 5 January 2018

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

(a)  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=0}^\infty \sqrt{n}x^n}

(b)  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=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}}

Background Information:  
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.


Solution:

(a)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
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{\sqrt{n+1}\cdot x^{n+1}}{\sqrt{n}\cdot x^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}\cdot x\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(\sqrt{\frac{n+1}{n}}\cdot|x|\bigg)}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{1}}\\ &&\\ &=& \displaystyle{|x|.} \end{array}}
Step 2:  
The Ratio Test tells us this 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 |x|<1.}
Hence, the Radius of Convergence of this series 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.}
Step 3:  
Now, we need to determine the interval of convergence.
First, 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 |x|<1}   corresponds to the interval  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 (-1,1).}
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  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.}
Step 4:  
First, 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=1.}
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=0}^\infty \sqrt{n}.}
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 \lim_{n\rightarrow \infty} \sqrt{n}=\infty.}
Therefore, the series diverges by the  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} th term test.
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=1}   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=-1.}
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=0}^\infty (-1)^n \sqrt{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 \lim_{n\rightarrow \infty} \sqrt{n}=\infty,}
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 \lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.}
Therefore, the series diverges by the  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} th term test.
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=-1 }   in the interval.
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 (-1,1).}

(b)

Step 1:  
We first use the Ratio Test to determine the radius of convergence.
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{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\cdot \frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)\cdot (x-3)\cdot \frac{2n+1}{2n+3}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg(|x-3|\cdot \frac{2n+1}{2n+3}\bigg)}\\ &&\\ & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\ &&\\ & = & \displaystyle{|x-3|.} \end{array}}

Step 2:  
The Ratio Test tells us this 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 |x-3|<1.}
Hence, the Radius of Convergence of this series 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.}
Step 3:  
Now, we need to determine the interval of convergence.
First, 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 |x-3|<1}   corresponds to the interval  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 (2,4).}
To obtain the interval of convergence, we need to test the endpoints of this interval
for convergence since the Ratio Test is inconclusive when  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.}
Step 4:  
First, 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=4.}
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=0}^\infty (-1)^n \frac{1}{2n+1}.}
This is an alternating 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 b_n=\frac{1}{2n+1}.} .
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}{2n+1}\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 0.}
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}{2(n+1)+1}<\frac{1}{2n+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.}
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}{2n+1}=0.}
Therefore, this series converges by the Alternating Series Test
and 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=4}   in our 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=0}^\infty \frac{1}{2n+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{1}{2n+1}>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 0.}
Thus, we can use the Limit Comparison Test.
We compare this series with 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},}
which is the harmonic series and divergent.
Now, 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} \frac{\big(\frac{1}{2n+1}\big)}{\big(\frac{1}{n}\big)}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\ &&\\ & = & \displaystyle{\frac{1}{2}.} \end{array}}

Since this limit is a finite number greater than zero,
       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=0}^\infty \frac{1}{2n+1}}  
diverges by the Limit Comparison Test.
Therefore, 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=2}   in our interval.
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 (2,4].}


Final Answer:  
    (a)     The radius of convergence is    and the interval of convergence is  
    (b)     The radius of convergence is    and the interval of convergence is  

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