Difference between revisions of "009C Sample Final 3, Problem 6"

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<span class="exam"> Consider the power series  
 
<span class="exam"> Consider the power series  
  
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::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}</math>
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::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
  
 
<span class="exam">(a) Find the radius of convergence of the above power series.
 
<span class="exam">(a) Find the radius of convergence of the above power series.
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::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>
 
::<math>\sum_{n=0}^\infty \frac{1}{(n+1)3^{n+1}}</math>
  
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<span class="exam">converge? If so, find its sum.
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<span class="exam">converge?
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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<hr>
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!Foundations: &nbsp;
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[[009C Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']]
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|-
 
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|'''1.''' '''Ratio Test'''
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math style="vertical-align: -7px">\sum a_n</math>&nbsp; be a series and &nbsp;<math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Then,
 
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|-
 
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|
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L<1,</math>&nbsp; the series is absolutely convergent.
 
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|-
 
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|
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L>1,</math>&nbsp; the series is divergent.
 
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|-
 
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|
 
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&nbsp; &nbsp; &nbsp; &nbsp; If &nbsp;<math style="vertical-align: -4px">L=1,</math>&nbsp; the test is inconclusive.
 
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|-
 
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|'''2.''' '''Direct Comparison Test'''
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; Let &nbsp;<math>\{a_n\}</math>&nbsp; and &nbsp;<math>\{b_n\}</math>&nbsp; be positive sequences where &nbsp;<math style="vertical-align: -3px">a_n\le b_n</math>
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; for all &nbsp;<math style="vertical-align: -3px">n\ge N</math>&nbsp; for some &nbsp;<math style="vertical-align: -3px">N\ge 1.</math>
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; '''1.''' If &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; converges, then &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; converges.
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; '''2.''' If &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; diverges, then &nbsp;<math>\sum_{n=1}^\infty b_n</math>&nbsp; diverges.
 
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|}
 
  
  
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'''Solution:'''
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[[009C Sample Final 3, Problem 6 Detailed Solution|'''<u>Detailed Solution</u>''']]
  
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'''(a)'''
 
  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|We use the Ratio Test to determine the radius of convergence.
 
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|-
 
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|We have
 
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|-
 
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|
 
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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)^{n+2}}{(n+2)}\frac{n+1}{(-1)^n(x)^{n+1}}\bigg|}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n+1}{n+2}\bigg|}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{n\rightarrow \infty} |x|\frac{n+1}{n+2}}\\
 
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&&\\
 
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& = & \displaystyle{|x|\lim_{n\rightarrow \infty} \frac{n+1}{n+2}}\\
 
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&&\\
 
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& = & \displaystyle{|x|.}
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|The Ratio Test tells us this series is absolutely convergent if &nbsp;<math style="vertical-align: -5px">|x|<1.</math>
 
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|-
 
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|Hence, the Radius of Convergence of this series is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
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|}
 
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'''(b)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|First, note that &nbsp;<math style="vertical-align: -5px">|x|<1</math>&nbsp; corresponds to the interval &nbsp;<math style="vertical-align: -4px">(-1,1).</math>
 
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|-
 
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|To obtain the interval of convergence, we need to test the endpoints of this interval
 
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|-
 
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|for convergence since the Ratio Test is inconclusive when &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
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|}
 
−
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|First, let &nbsp;<math style="vertical-align: -1px">x=1.</math> 
 
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|-
 
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|Then, the series becomes &nbsp;<math>\sum_{n=0}^\infty (-1)^n \frac{1}{n+1}.</math>
 
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|-
 
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|This is an alternating series.
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
 
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|-
 
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|The sequence &nbsp;<math>\{b_n\}</math>&nbsp; is decreasing since
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
 
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|-
 
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
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|-
 
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|Also,
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{n+1}=0.</math>
 
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|-
 
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|Therefore, this series converges by the Alternating Series Test
 
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|-
 
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|and we include &nbsp;<math style="vertical-align: -1px">x=1</math>&nbsp; in our interval.
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Now, let &nbsp;<math style="vertical-align: -1px">x=-1.</math>
 
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|-
 
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|Then, the series becomes
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\sum_{n=0}^\infty \frac{(-1)^{2n+1}}{n+1}} & = & \displaystyle{\sum_{n=1}^\infty \frac{-1}{n+1}}\\
 
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&&\\
 
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& = & \displaystyle{(-1)\sum_{n=1}^\infty \frac{1}{n+1}.}
 
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\end{array}</math>
 
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|-
 
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|Now, we note that
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{n+1}>0</math>
 
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|-
 
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
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|-
 
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|This means that we can use the limit comparison test on this series.
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -16px">a_n=\frac{1}{n+1}.</math>
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
 
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|-
 
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|Then, &nbsp;<math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>&nbsp; diverges since it is the harmonic series.
 
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|-
 
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|We have
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{\lim_{n\rightarrow \infty} \frac{a_n}{b_n}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{(\frac{1}{n+1})}{(\frac{1}{n})}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{n+1}}\\
 
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&&\\
 
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& = & \displaystyle{\lim_{n\rightarrow \infty} 1.}
 
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\end{array}</math>
 
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|-
 
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|Therefore, the series
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^{\infty} \frac{1}{n+1}</math>
 
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|-
 
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|diverges by the Limit Comparison Test.
 
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|-
 
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|Therefore, we do not include &nbsp;<math style="vertical-align: -1px">x=-1</math>&nbsp; in our interval.
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 4: &nbsp;
 
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|-
 
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|The interval of convergence is &nbsp;<math style="vertical-align: -4px">(-1,1].</math>
 
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|}
 
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'''(c)'''
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 1: &nbsp;
 
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|-
 
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|Let
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp;<math>f(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math>
 
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|-
 
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|Then,
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f'(x)} & = & \displaystyle{\frac{d}{dx}\bigg(\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{\sum_{n=0}^\infty \frac{d}{dx}\bigg( (-1)^n \frac{x^{n+1}}{n+1}\bigg)}\\
 
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&&\\
 
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& = & \displaystyle{\sum_{n=0}^\infty (-1)^n x^n}\\
 
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&&\\
 
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& = & \displaystyle{\sum_{n=0}^\infty (-x)^n}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{1-(-x)}}\\
 
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&&\\
 
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& = & \displaystyle{\frac{1}{1+x}.}
 
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\end{array}</math>
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Then,
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
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\displaystyle{f(x)} & = & \displaystyle{\int \frac{1}{1+x}~dx}\\
 
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&&\\
 
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& = & \displaystyle{\ln(1+x)+C.}
 
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\end{array}</math>
 
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|-
 
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|Since there is no constant term in the series <math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>C=0.</math>
 
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|-
 
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|Hence,
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>f(x)=\ln(1+x).</math>
 
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|}
 
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'''(d)'''
 
−
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
−
!Step 1: &nbsp;
 
−
|-
 
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|First, we note that
 
−
|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)3^{n+1}}>0</math>
 
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|-
 
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
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|-
 
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|This means that we can use a comparison test on this series.
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -19px">a_n=\frac{1}{(n+1)3^{n+1}}.</math>
 
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|}
 
−
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 2: &nbsp;
 
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|-
 
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|Let &nbsp;<math style="vertical-align: -14px">b_n=\frac{1}{3^{n+1}}.</math>
 
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|-
 
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|We want to compare the series in this problem with
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty b_n=\sum_{n=1}^\infty \frac{1}{3}\bigg(\frac{1}{3}\bigg)^n.</math>
 
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|-
 
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|This is a geometric series with &nbsp;<math style="vertical-align: -13px">r=\frac{1}{3}.</math>
 
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|-
 
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|Since &nbsp; <math>|r|<1,</math>&nbsp; the series &nbsp;<math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math>&nbsp; converges.
 
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|}
 
−
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Step 3: &nbsp;
 
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|-
 
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|Also, we have &nbsp;<math style="vertical-align: -4px">a_n<b_n</math>&nbsp; since
 
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|-
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{(n+1)3^{n+1}}<\frac{1}{3^{n+1}}</math>
 
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|-
 
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|for all &nbsp;<math style="vertical-align: -3px">n\ge 0.</math>
 
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|-
 
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|Therefore, the series &nbsp;<math>\sum_{n=1}^\infty a_n</math>&nbsp; converges
 
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|-
 
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|by the Direct Comparison Test.
 
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|}
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
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!Final Answer: &nbsp;
 
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|-
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; The radius of convergence is &nbsp;<math style="vertical-align: -1px">R=1.</math>
 
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|-
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>(-1,1]</math>
 
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|-
 
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|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp; <math>f(x)=\ln(1+x)</math>
 
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|-
 
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|&nbsp; &nbsp; '''(d)''' &nbsp; &nbsp; converges
 
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|}
 
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 15:21, 3 December 2017

Consider the power 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=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.}

(a) Find the radius of convergence of the above power series.

(b) Find the interval of convergence of the above power series.

(c) Find the closed formula for the function  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 f(x)}   to which the power series converges.

(d) Does 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=0}^\infty \frac{1}{(n+1)3^{n+1}}}

converge?

Solution


Detailed Solution


Return to Sample Exam

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