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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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| | + | [[009C Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !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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| − | '''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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| − | '''(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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n}
of the series.
(b) Compute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty} s_n.}
Solution
Detailed Solution
Return to Sample Exam