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| | <span class="exam"> Find the radius of convergence and interval of convergence of the series. | | <span class="exam"> Find the radius of convergence and interval of convergence of the series. |
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| − | ::<span class="exam">a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math>
| + | <span class="exam">(a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> |
| − | ::<span class="exam">b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math>
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <span class="exam">(b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math> |
| − | !Foundations:
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| − | |-
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| − | |Ratio Test
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| − | ::
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| − | ::
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| − | |}
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| | + | <hr> |
| | + | [[009C Sample Midterm 1, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)''' | + | [[009C Sample Midterm 1, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |We first 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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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty}|\frac{a_{n+1}}{a_n}|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|\sqrt{1}}\\
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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:
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| − | |-
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| − | |The Ratio Test tells us this series is absolutely convergent if <math>|x|<1.</math>
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| − | |-
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| − | |Hence, the Radius of Convergence of this series is <math>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 3:
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| − | |-
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| − | |Now, we need to determine the interval of convergence.
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| − | |-
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| − | |First, note that <math>|x|<1</math> corresponds to the interval <math>(-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>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 4:
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| − | |-
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| − | |First, let <math>x=1.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math>
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| − | |-
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| − | |We note that
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| − | |-
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| − | | <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math>
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| − | |Therefore, the series diverges by the <math>n</math>th term test.
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| − | |-
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| − | |Hence, we do not include <math>x=1</math> in the 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 5:
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| − | |-
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| − | |Now, let <math>x=-1.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math>
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| − | |-
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| − | |Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math>
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| − | |-
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| − | |we have
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| − | |-
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| − | | <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=</math>DNE.
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| − | |-
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| − | |Therefore, the series diverges by the <math>n</math>th term test.
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| − | |-
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| − | |Hence, we do not include <math>x=-1 </math> in the 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 6:
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| − | |-
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| − | |The interval of convergence is <math>(-1,1).</math>
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| − | |}
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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:
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| − | |-
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| − | |We first 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-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\
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| − | &&\\
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| − | & = & \displaystyle{|x-3|}
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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>|x-3|<1.</math>
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| − | |-
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| − | |Hence, the Radius of Convergence of this series is <math>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 3:
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| − | |-
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| − | |Now, we need to determine the interval of convergence.
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| − | |-
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| − | |First, note that <math>|x-3|<1</math> corresponds to the interval <math>(2,4).</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>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 4:
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| − | |-
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| − | |First, let <math>x=4.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+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>b_n=\frac{1}{2n+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}{2(n+1)+1}<\frac{1}{2n+1}</math>
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| − | |-
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| − | |for all <math>n\ge 1.</math>
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| − | |-
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| − | |Also,
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| − | | <math>\lim_{n\rightarrow \infty} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.</math>
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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>x=4</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 5:
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| − | |-
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| − | |Now, let <math>x=2.</math>
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| − | |-
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| − | |Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math>
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| − | |-
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| − | |First, we note that <math>\frac{1}{2n+1}>0</math> for all <math>n\ge 0.</math>
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| − | |-
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| − | |Thus, we can use the Limit Comparison Test.
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| − | |-
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| − | |We compare this series with the series <math>\sum_{n=1}^\infty \frac{1}{n},</math>
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| − | |-
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| − | |which is the harmonic series and divergent.
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| − | |-
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| − | |Now, 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} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}.}
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| − | \end{array}</math>
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| − | |-
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| − | |Since this limit is a finite number greater than zero, we have
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| − | |-
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| − | |<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> diverges by the
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| − | |-
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| − | |Limit Comparison Test. Therefore, we do not include <math>x=2</math>
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| − | |-
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| − | |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 6:
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| − | |-
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| − | |The interval of convergence is <math>(2,4].</math>
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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>R=1</math> and the interval of convergence is <math>(-1,1).</math>
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| − | |-
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| − | | '''(b)''' The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>(2,4].</math>
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| − | |}
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| | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
