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| | <span class="exam">(b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math> | | <span class="exam">(b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math> |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009C Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | '''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
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| − | <math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
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| − | '''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
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| − | and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
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| − | '''Solution:''' | + | [[009C Sample Final 1, 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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| − | |First, we write
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\
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| − | &&\\
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| − | & = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n.}\\
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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 <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math>
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| − | |So,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\sum_{n=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{1}{\frac{e+2}{e}}}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{e}{e+2}.}\\
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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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| − | |This is a telescoping series. First, we find the partial sum of this series.
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| − | |Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
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| − | |Then,
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| − | <math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+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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| − | |Thus,
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{\sum_{n=1}^{\infty}\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} & = & \displaystyle{\lim_{k\rightarrow \infty} s_k}\\
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| − | &&\\
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| − | & = & \displaystyle{\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+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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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' <math>\frac{e}{e+2}</math>
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| − | | '''(b)''' <math>\frac{1}{2}</math>
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| − | |}
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| | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
