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| − | <span class="exam">Find the radius of convergence and the interval of convergence | + | <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) <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}.</math>
| + | <span class="exam">(a) <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(-1)^{n}x^{n}}{n+1}</math> |
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| − | ::<span class="exam">(b) (6 points) <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}.</math>
| + | <span class="exam">(b) <math>{\displaystyle \sum_{n=0}^{\infty}}\frac{(x+1)^{n}}{n^{2}}</math> |
| | + | <hr> |
| | + | [[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:
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| − | |-
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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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| − | |}
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| − | '''Solution:'''
| + | [[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):
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| − | |-
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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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| − | |
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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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| − | |
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| − | ::<math>a_{n}\,=\,\frac{(-1)^{n}x^{n}}{n+1}\,=\,\frac{(-1)^{n}}{n+1},</math>
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| − | |-
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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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| − | |
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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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| − | |-
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(b):
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| − | |-
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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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| − | |-
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| − | |
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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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| − | |-
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| − | |In this case, the radius is 1, and the interval will be centered at <math style="vertical-align: -2%">x=-1</math>, or when <math style="vertical-align: -5%">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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| − | |
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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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| − | |-
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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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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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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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| − | |-
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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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| − | |}
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| | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
