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| | <span class="exam">(b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math> | | <span class="exam">(b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math> |
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| | + | (insert picture of handwritten solution) |
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| | + | [[009C Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution for this Problem</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |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:'''
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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 need to find a pattern for the partial sums in order to find a formula.
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| − | |We start by calculating <math style="vertical-align: -3px">s_2.</math> We have
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| − | | <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</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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| − | |Next, we calculate <math style="vertical-align: -3px">s_3</math> and <math style="vertical-align: -3px">s_4.</math> We have
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{s_3} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\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_4} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)+2\bigg(\frac{1}{2^4}-\frac{1}{2^5}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).}
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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 3:
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| − | |If we look at <math style="vertical-align: -4px">s_2,s_3,</math> and <math style="vertical-align: -4px">s_4, </math> we notice a pattern.
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| − | |From this pattern, we get the formula
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| − | | <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</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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| − | |From Part (a), we have
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| − | | <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+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 2:
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| − | |We now calculate <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
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| − | |We get
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| − | | <math>\begin{array}{rcl}
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| − | \displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} 2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{2}{2^2}}\\
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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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)</math>
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| − | | '''(b)''' <math>\frac{1}{2}</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>''']] |
Consider the infinite series 
(a) Find an expression for the 
th partial sum 
of the series.
(b) Compute 
(insert picture of handwritten solution)
Detailed Solution for this Problem
Return to Sample Exam