Difference between revisions of "009C Sample Midterm 3, Problem 5"

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<span class="exam">Find the radius of convergence and the interval of convergence
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<span class="exam"> Find the radius of convergence and the interval of convergence
 
of the series.  
 
of the series.  
  
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::<span class="exam">(a) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}.</math>
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<span class="exam">(a) &nbsp;<math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}</math>
  
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::<span class="exam">(b) (6 points) &nbsp;&nbsp;&nbsp;&nbsp; <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}.</math>
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<span class="exam">(b) &nbsp;<math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}</math>
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<hr>
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[[009C Sample Midterm 3, Problem 5 Solution|'''<u>Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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! Foundations: &nbsp;
 
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|When we are asked to find the radius of convergence, we are given a series where
 
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::<math>a_n=f(x-c)\cdot g(n)</math>
 
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|where <math style="vertical-align: -20%">f</math> and <math style="vertical-align: -20%">g</math> are functions of <math style="vertical-align: 0%">x</math> and <math style="vertical-align: 0%">n</math> respectively, and <math style="vertical-align: 0%">c</math> is a constant (frequently zero).  We need to find a bound (radius) on <math style="vertical-align: -22%">|x-c|</math> such that whenever <math style="vertical-align: -22%">|x-c|<r</math>, the ratio test
 
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::<math>\left|\frac{a_{n+1}}{a_n}\right|<1.</math>
 
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When we do, the interval will be <math style="vertical-align: -20%">(c-r,c+r)</math>.  However, the boundary values for <math style="vertical-align: 0%">x</math>, <math style="vertical-align: 0%">c-r</math> and <math style="vertical-align: -8%">c+r</math> must be tested individually for convergence.  Many times, one boundary value will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence may open, half-open or closed.
 
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&nbsp;'''Solution:'''
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[[009C Sample Midterm 3, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(a): &nbsp;
 
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|We need to choose a radius <math style="vertical-align: 0%">r</math> such that whenever <math style="vertical-align: -20%">|x|<r</math>,
 
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::<math>\begin{array}{rcl}
 
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\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}x^{n+1}}{n+2}}{\frac{(-1)^{n}x^{n}}{n+1}}\right|\\
 
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\\
 
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& = & \lim_{n\rightarrow\infty}\left|x\cdot\frac{n+1}{n+2}\right|\\
 
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\\
 
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& = & |x|\cdot\lim_{n\rightarrow\infty}\left|\frac{n+1}{n+2}\right|\\
 
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\\
 
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& = & |x|<1.
 
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\end{array}</math>
 
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|In this case, the radius is 1, and the interval will be centered at 0. We then need to take a look at the boundary points. If <math style="vertical-align: -20%">x=1,</math> then
 
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::<math>a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}}{n+1},</math>
 
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|so the series is an alternating harmonic series which converges. On the other hand, if <math style="vertical-align: -20%">x=-1,</math> then
 
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::<math>a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}(-1)^{n}}{n+1}\,=\,\frac{1}{n+1},</math>
 
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|a standard harmonic series which does not converge. Thus, the interval of convergence is <math style="vertical-align: -20%">(-1,1]</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!(b): &nbsp;
 
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|We need to choose a radius <math style="vertical-align: 0%">r</math> such that whenever <math style="vertical-align: -20%">|x|<r</math>,
 
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::<math>\begin{array}{rcl}
 
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\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right| & = & \lim_{n\rightarrow\infty}\left|\frac{\frac{(x+1)^{n+1}}{(n+1)^{2}}}{\frac{(x+1)^{n}}{n^{2}}}\right|\\
 
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\\
 
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& = & \lim_{n\rightarrow\infty}\left|(x+1)\cdot\left(\frac{n}{n+1}\right)^{2}\right|\\
 
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\\
 
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& = & |x+1|\cdot\lim_{n\rightarrow\infty}\left(\frac{n}{n+1}\right)^{2}\\
 
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\\
 
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& = & |x+1|<1.
 
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\end{array}</math>
 
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|In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: 0%">x=-1</math>, or when <math style="vertical-align: -10%">x+1=0</math>. We then need to take a look at the boundary points. If <math style="vertical-align: 0%">x=-2</math> or <math style="vertical-align: 0%">x=0</math>, then
 
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::<math>\left|a_{n}\right|\,=\,\left|\frac{(x+1)^{n}}{n^{2}}\right|\,=\,\frac{1}{n^{2}},</math>
 
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|which defines a p-series with <math style="vertical-align: -20%">p=2</math>. Thus, the series defined by each boundary point is absolutely convergent (and therefore convergent), and the interval of convergence is <math style="vertical-align: -20%">[-2,0]</math>.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|For (a), the radius is 1 and the interval of convergence is <math style="vertical-align: -20%">(-1,1]</math>.
 
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|For (b), the radius is also 1, but the interval of convergence is <math style="vertical-align: -20%">[-2,0]</math>.
 
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[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 09:47, 28 November 2017

Find the radius of convergence and the 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 {\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}}

(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 {\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}}

Solution


Detailed Solution


Return to Sample Exam

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