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>
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<span class="exam">Consider the infinite series &nbsp;<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 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">(a) Find an expression for the &nbsp;<math style="vertical-align: 0px">n</math>th partial sum &nbsp;<math style="vertical-align: -3px">s_n</math>&nbsp; of the series.
  
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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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<span class="exam">(b) Compute &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>
  
  
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!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
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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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|The &nbsp;<math style="vertical-align: 0px">n</math>th partial sum, &nbsp;<math style="vertical-align: -3px">s_n</math>&nbsp; for a series &nbsp;<math>\sum_{n=1}^\infty a_n </math>&nbsp; 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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|We start by calculating <math style="vertical-align: -3px">s_2</math>. We have
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|We start by calculating &nbsp;<math style="vertical-align: -3px">s_2.</math>&nbsp; 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 style="vertical-align: -3px">s_3</math> and <math style="vertical-align: -3px">s_4.</math> We have
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|Next, we calculate &nbsp;<math style="vertical-align: -3px">s_3</math>&nbsp; and &nbsp;<math style="vertical-align: -3px">s_4.</math>&nbsp; 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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|-
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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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|If we look at &nbsp;<math style="vertical-align: -4px">s_2,s_3,</math>&nbsp; and &nbsp;<math style="vertical-align: -4px">s_4, </math>&nbsp; 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 style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>  
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|We now calculate &nbsp;<math style="vertical-align: -11px">\lim_{n\rightarrow \infty} s_n.</math>  
 
|-
 
|-
 
|We get  
 
|We get  

Revision as of 18:50, 26 February 2017

Consider the infinite series  

(a) Find an expression for the  th partial sum    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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