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| | <span class="exam">(a) <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math> | | <span class="exam">(a) <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math> |
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| − | <span class="exam">(b) <math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math> | + | <span class="exam">(b) <math>\sum_{n=1}^{\infty} \frac{1}{(2n-1)(2n+1)}</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 2, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | |'''1.''' The sum of a convergent geometric series is <math>\frac{a}{1-r}</math> | |
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| − | | where <math style="vertical-align: 0px">r</math> is the ratio of the geometric series
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| − | | and <math style="vertical-align: 0px">a</math> is the first term of the series.
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| − | |'''2.''' The <math style="vertical-align: 0px">n</math>th partial sum, <math style="vertical-align: -3px">s_n</math> for a series <math>\sum_{n=1}^\infty a_n </math> is defined as
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| − | <math>s_n=\sum_{i=1}^n a_i.</math>
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| − | |}
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| − | '''Solution:''' | + | [[009C Sample Final 2, Problem 2 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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| − | |Let <math style="vertical-align: -3px">a_n</math> be the <math style="vertical-align: 0px">n</math>th term of this sum.
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| − | |We notice that
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| − | | <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math> and <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
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| − | |So, this is a geometric series with <math style="vertical-align: -14px">r=-\frac{1}{2}.</math>
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| − | |Since <math style="vertical-align: -5px">|r|<1,</math> this series converges.
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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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| − | |Hence, the sum of this geometric series is
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{8}{3}.}
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| − | \end{array}</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:
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| − | |We begin by using partial fraction decomposition. Let
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| − | | <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math>
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| − | |If we multiply this equation by <math style="vertical-align: -5px">(2x-1)(2x+1),</math> we get
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| − | | <math>1=A(2x+1)+B(2x-1).</math>
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| − | |If we let <math style="vertical-align: -14px">x=\frac{1}{2},</math> we get <math style="vertical-align: -14px">A=\frac{1}{2}.</math>
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| − | |If we let <math style="vertical-align: -14px">x=-\frac{1}{2},</math> we get <math style="vertical-align: -14px">B=-\frac{1}{2}.</math>
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| − | |So, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{-\frac{1}{2}}{2n+1}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.}
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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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| − | |Now, we look at the partial sums, <math>s_n</math> of this series.
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| − | |First, we have
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| − | | <math>s_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).</math>
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| − | |Also, we have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{s_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\bigg)}
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| − | \end{array}</math>
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| − | |and
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{s_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).}
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| − | \end{array}</math>
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| − | |If we compare <math>s_1,s_2,s_3,</math> we notice a pattern.
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| − | |We have
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| − | | <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</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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| − | |Now, to calculate the sum of this series we need to calculate
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| − | | <math>\lim_{n\rightarrow \infty} s_n.</math>
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| − | |We have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{2}.}
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| − | \end{array}</math>
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| − | |Since the partial sums converge, the series converges and the sum of the series is <math style="vertical-align: -15px">\frac{1}{2}.</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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| − | | '''(a)''' <math>\frac{8}{3}</math>
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| − | | '''(b)''' <math>\frac{1}{2}</math>
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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>''']] |
