|
|
| (One intermediate revision by the same user not shown) |
| Line 4: |
Line 4: |
| | | | |
| | <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> |
| | + | <hr> |
| | + | [[009C Sample Midterm 1, Problem 2 Solution|'''<u>Solution</u>''']] |
| | | | |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !Foundations:
| |
| − | |-
| |
| − | |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
| |
| − | |- | |
| − | |
| |
| − | <math>s_n=\sum_{i=1}^n a_i.</math>
| |
| − | |}
| |
| | | | |
| | | | |
| − | '''Solution:'''
| |
| − |
| |
| − | '''(a)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |We need to find a pattern for the partial sums in order to find a formula.
| |
| − | |-
| |
| − | |We start by calculating <math style="vertical-align: -3px">s_2.</math> We have
| |
| − | |-
| |
| − | | <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Next, we calculate <math style="vertical-align: -3px">s_3</math> and <math style="vertical-align: -3px">s_4.</math> We have
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \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)}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\bigg)}
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |and
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \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)}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 3:
| |
| − | |-
| |
| − | |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.
| |
| − | |-
| |
| − | |From this pattern, we get the formula
| |
| − | |-
| |
| − | | <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |From Part (a), we have
| |
| − | |-
| |
| − | | <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |We now calculate <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
| |
| − | |-
| |
| − | |We get
| |
| − | |-
| |
| − | | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} 2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{2}{2^2}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{1}{2}.}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | | '''(a)''' <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)</math>
| |
| − | |-
| |
| − | | '''(b)''' <math>\frac{1}{2}</math>
| |
| − | |}
| |
| | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] |
