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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> | + | ::<math>\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1}.</math> |
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| | <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. | + | <span class="exam">converge? |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 3, Problem 6 Solution|'''<u>Solution</u>''']] |
| − | |-
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| − | |'''1.''' '''Ratio Test'''
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| − | |-
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| − | | Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math>
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| − | |-
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| − | | Then,
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| − | |-
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| − | |
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| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
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| − | |-
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| − | |
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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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| − | |-
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| − | |
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| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
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| − | |- | |
| − | |'''2.''' '''Direct Comparison Test'''
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| − | |-
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| − | | Let <math>\{a_n\}</math> and <math>\{b_n\}</math> be positive sequences where <math style="vertical-align: -3px">a_n\le b_n</math>
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| − | |-
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| − | | for all <math style="vertical-align: -3px">n\ge N</math> for some <math style="vertical-align: -3px">N\ge 1.</math>
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| − | |-
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| − | | '''1.''' If <math>\sum_{n=1}^\infty b_n</math> converges, then <math>\sum_{n=1}^\infty a_n</math> converges.
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| − | |-
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| − | | '''2.''' If <math>\sum_{n=1}^\infty a_n</math> diverges, then <math>\sum_{n=1}^\infty b_n</math> diverges.
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| − | |}
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| − | '''Solution:''' | + | [[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:
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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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| − | <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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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |The Ratio Test tells us this series is absolutely convergent if <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 <math style="vertical-align: -1px">R=1.</math>
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| − | |}
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| − |
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| − | '''(b)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, note that <math style="vertical-align: -5px">|x|<1</math> corresponds to the interval <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 <math style="vertical-align: -1px">R=1.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |First, let <math style="vertical-align: -1px">x=1.</math>
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| − | |-
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| − | |Then, the series becomes <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 <math style="vertical-align: -15px">b_n=\frac{1}{n+1}.</math>.
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| − | |-
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| − | |The sequence <math>\{b_n\}</math> is decreasing since
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| − | |-
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| − | | <math>\frac{1}{n+2}<\frac{1}{n+1}</math>
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| − | |-
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| − | |for all <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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| − | | <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 <math style="vertical-align: -1px">x=1</math> in our interval.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Now, let <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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| − | | <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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| − | | <math>\frac{1}{n+1}>0</math>
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| − | |-
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| − | |for all <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 <math style="vertical-align: -16px">a_n=\frac{1}{n+1}.</math>
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| − | |-
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| − | |Let <math style="vertical-align: -14px">b_n=\frac{1}{n}.</math>
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| − | |-
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| − | |Then, <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math> 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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| − | | <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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| − | | <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 <math style="vertical-align: -1px">x=-1</math> in our interval.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 4:
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| − | |-
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| − | |The interval of convergence is <math style="vertical-align: -4px">(-1,1].</math>
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| − | |}
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| − |
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| − | '''(c)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |Let
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| − | |-
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| − | | <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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| − | | <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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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Thus,
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| − | |-
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| − | | <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 style="vertical-align: -20px">\sum_{n=0}^\infty (-1)^n \frac{x^{n+1}}{n+1},</math>
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| − | |-
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| − | | <math>C=0.</math>
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| − | |-
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| − | |Hence,
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| − | |-
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| − | | <math>f(x)=\ln(1+x).</math>
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| − | |}
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| − |
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| − | '''(d)'''
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we note that
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| − | |-
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| − | | <math>\frac{1}{(n+1)3^{n+1}}>0</math>
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| − | |-
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| − | |for all <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 <math style="vertical-align: -19px">a_n=\frac{1}{(n+1)3^{n+1}}.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |Let <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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| − | | <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 <math style="vertical-align: -13px">r=\frac{1}{3}.</math>
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| − | |-
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| − | |Since <math>|r|<1,</math> the series <math style="vertical-align: -20px">\sum_{n=1}^\infty b_n</math> converges.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 3:
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| − | |-
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| − | |Also, we have <math style="vertical-align: -4px">a_n<b_n</math> since
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| − | |-
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| − | | <math>\frac{1}{(n+1)3^{n+1}}<\frac{1}{3^{n+1}}</math>
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| − | |-
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| − | |for all <math style="vertical-align: -3px">n\ge 0.</math>
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| − | |-
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| − | |Therefore, the series <math>\sum_{n=1}^\infty a_n</math> converges
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| − | |-
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| − | |by the Direct Comparison Test.
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| − | |}
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| − |
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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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| − | |-
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| − | | '''(a)''' The radius of convergence is <math style="vertical-align: -1px">R=1.</math>
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| − | |-
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| − | | '''(b)''' <math>(-1,1]</math>
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| − | |-
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| − | | '''(c)''' <math>f(x)=\ln(1+x)</math>
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| − | |-
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| − | | '''(d)''' converges
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| − | |}
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| | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] |
