MathAdmin at 05:54, 12 November 2017

2017-11-12T05:54:53Z

MathAdmin: Created page with "Consider the infinite series $\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).$ (a) Find an expression f..."

2017-04-09T19:26:33Z

Created page with "Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math> (a) Find an expression f..."

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Consider the infinite series  <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>

(a) Find an expression for the  <math style="vertical-align: 0px">n</math>th partial sum  <math style="vertical-align: -3px">s_n</math>  of the series.

(b) Compute  <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!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|'''Return to Sample Exam''']]

@@ Line 4: / Line 4: @@
 <span class="exam">(b) Compute &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+[[009C Sample Midterm 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]

009C Sample Midterm 1, Problem 2 - Revision history

MathAdmin at 05:54, 12 November 2017

MathAdmin: Created page with "Consider the infinite series \sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg). (a) Find an expression f..."

MathAdmin: Created page with "Consider the infinite series $\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).$ (a) Find an expression f..."