Difference between revisions of "009C Sample Midterm 1, Problem 2"

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<span class="exam">Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
 
<span class="exam">Consider the infinite series <math>\sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
  
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::<span class="exam">a) Find an expression for the <math>n</math>th partial sum <math>s_n</math> of the series.
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::<span class="exam">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.
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::<span class="exam">b) Compute <math>\lim_{n\rightarrow \infty} s_n.</math>
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::<span class="exam">b) Compute <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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|-
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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>  
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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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|-
 
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|is defined as
 
 
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|-
 
|
 
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|We need to find a pattern for the partial sums in order to find a formula.  
 
|We need to find a pattern for the partial sums in order to find a formula.  
 
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|-
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|We start by calculating <math>s_2</math>. We have
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|We start by calculating <math style="vertical-align: -3px">s_2</math>. We have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).</math>
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|-
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|Next, we calculate <math>s_3</math> and <math>s_4.</math> We have
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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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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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!Step 3: &nbsp;  
 
!Step 3: &nbsp;  
 
|-
 
|-
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|If we look at <math>s_2,s_3,s_4, </math> we notice a pattern.
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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
 
|From this pattern, we get the formula
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
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|We now calculate <math>\lim_{n\rightarrow \infty} s_n.</math>  
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|We now calculate <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>  
 
|-
 
|-
 
|We get  
 
|We get  

Revision as of 08:36, 14 February 2017

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.}


Foundations:  
The 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 n} 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} for a series 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 \sum_{n=1}^\infty a_n } is defined as

        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=\sum_{i=1}^n a_i.}


Solution:

(a)

Step 1:  
We need to find a pattern for the partial sums in order to find a formula.
We start by calculating 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_2} . We have
        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_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).}
Step 2:  
Next, we calculate 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_3} and 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_4.} We have
        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 \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}}
and
        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 \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}}
Step 3:  
If we look at 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_2,s_3,} and 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_4, } we notice a pattern.
From this pattern, we get the formula
        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=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).}

(b)

Step 1:  
From Part (a), we have
        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=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).}
Step 2:  
We now calculate 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.}
We get
        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 \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}}


Final Answer:  
    (a)     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=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg)}
    (b)     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 \frac{1}{2}}

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