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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|-
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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  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=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).}

(a) Find an expression for 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}   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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