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| | <span class="exam">(b) <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math> | | <span class="exam">(b) <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 3 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' '''Ratio Test'''
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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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| − | | Then,
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| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent.
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| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent.
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| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive.
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| − | |'''2.''' If a series absolutely converges, then it also converges.
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| − | |'''3.''' '''Alternating Series Test''' | |
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| − | | Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math>
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| − | | Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math>
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| − | | converge.
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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 3 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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| − | |We begin by using the Ratio Test.
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| − | |We have
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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{(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \frac{(2n)!}{n!}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\
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| − | &&\\
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| − | & = & \displaystyle{0.}
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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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| − | |Since
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| − | | <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=0<1,</math>
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| − | |the series is absolutely convergent by the Ratio Test.
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| − | |Therefore, the series converges.
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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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| − | |For
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| − | | <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math>
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| − | |we notice that this series is alternating.
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| − | |Let <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math>
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| − | |First, we have
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| − | | <math>\frac{1}{n+1}\ge 0</math>
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| − | |for all <math style="vertical-align: -3px">n\ge 1.</math>
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| − | |The sequence <math style="vertical-align: -5px">\{b_n\}</math> is decreasing since
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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 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:
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| − | |Also,
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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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| − | |Therefore, the series <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math> converges
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| − | |by the Alternating Series Test.
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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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| − | | '''(a)''' converges (by the Ratio Test)
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| − | | '''(b)''' converges (by the Alternating Series Test)
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| − | |}
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| | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
