Difference between revisions of "009C Sample Final 2, Problem 2"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 28: | Line 28: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | |Let <math>a_n</math> be the <math>n</math>th term of this sum. | + | |Let <math style="vertical-align: -3px">a_n</math> be the <math style="vertical-align: 0px">n</math>th term of this sum. |
|- | |- | ||
|We notice that | |We notice that | ||
|- | |- | ||
| − | |   | + | | <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math> and <math>\frac{a_4}{a_2}=\frac{-1}{2}.</math> |
|- | |- | ||
| − | |So, this is a geometric series with <math>r=\frac{-1}{2}.</math> | + | |So, this is a geometric series with <math style="vertical-align: -14px">r=\frac{-1}{2}.</math> |
|- | |- | ||
| − | |Since <math>|r|<1,</math> this series converges. | + | |Since <math style="vertical-align: -5px">|r|<1,</math> this series converges. |
|} | |} | ||
| Line 63: | Line 63: | ||
| <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math> | | <math>\frac{1}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.</math> | ||
|- | |- | ||
| − | |If we multiply this equation by <math>(2x-1)(2x+1),</math> we get | + | |If we multiply this equation by <math style="vertical-align: -5px">(2x-1)(2x+1),</math> we get |
|- | |- | ||
| <math>1=A(2x+1)+B(2x-1).</math> | | <math>1=A(2x+1)+B(2x-1).</math> | ||
|- | |- | ||
| − | |If we let <math>x=\frac{1}{2},</math> we get <math>A=\frac{1}{2}.</math> | + | |If we let <math style="vertical-align: -14px">x=\frac{1}{2},</math> we get <math style="vertical-align: -14px">A=\frac{1}{2}.</math> |
|- | |- | ||
| − | |If we let <math>x=\frac{-1}{2},</math> we get <math>B=\frac{-1}{2}.</math> | + | |If we let <math style="vertical-align: -14px">x=\frac{-1}{2},</math> we get <math style="vertical-align: -14px">B=\frac{-1}{2}.</math> |
|- | |- | ||
|So, we have | |So, we have | ||
| Line 107: | Line 107: | ||
|If we compare <math>s_1,s_2,s_3,</math> we notice a pattern. | |If we compare <math>s_1,s_2,s_3,</math> we notice a pattern. | ||
|- | |- | ||
| − | |We have <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math> | + | |We have |
| + | |- | ||
| + | | <math>s_n=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).</math> | ||
|} | |} | ||
| Line 125: | Line 127: | ||
\end{array}</math> | \end{array}</math> | ||
|- | |- | ||
| − | |Since the partial sums converge, the series converges and the sum of the series is <math>\frac{1}{2}.</math> | + | |Since the partial sums converge, the series converges and the sum of the series is <math style="vertical-align: -15px">\frac{1}{2}.</math> |
|} | |} | ||
Revision as of 19:06, 4 March 2017
For each of the following series, find the sum if it converges. If it diverges, explain why.
(a)
(b)
| Foundations: |
|---|
| 1. The sum of a convergent geometric series is |
| where is the ratio of the geometric series |
| and is the first term of the series. |
| 2. The th partial sum, for a series is defined as |
|
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle s_{n}=\sum _{i=1}^{n}a_{i}.} |
Solution:
(a)
| Step 1: |
|---|
| Let be the th term of this sum. |
| We notice that |
| and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {a_{4}}{a_{2}}}={\frac {-1}{2}}.} |
| So, this is a geometric series with Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle r={\frac {-1}{2}}.} |
| Since this series converges. |
| Step 2: |
|---|
| Hence, the sum of this geometric series is |
|
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\frac {a_{1}}{1-r}}&=&\displaystyle {\frac {4}{1-(-{\frac {1}{2}})}}\\&&\\&=&\displaystyle {\frac {4}{\frac {3}{2}}}\\&&\\&=&\displaystyle {{\frac {8}{3}}.}\end{array}}} |
(b)
| Step 1: |
|---|
| We begin by using partial fraction decomposition. Let |
| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {1}{(2x-1)(2x+1)}}={\frac {A}{2x-1}}+{\frac {B}{2x+1}}.} |
| If we multiply this equation by we get |
| If we let 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 x=\frac{1}{2},} 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 A=\frac{1}{2}.} |
| If we let 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 x=\frac{-1}{2},} 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 B=\frac{-1}{2}.} |
| So, 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{\sum_{n=1}^\infty \frac{1}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{1}{2}}{2n-1}+\frac{\frac{-1}{2}}{2n+1}}\\ &&\\ & = & \displaystyle{\frac{1}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.} \end{array}} |
| Step 2: |
|---|
| Now, we look at the partial sums, 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 this series. |
| First, 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_1=\frac{1}{2}\bigg(1-\frac{1}{3}\bigg).} |
| Also, 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_2} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\ &&\\ & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{5}\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_3} & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\ &&\\ & = & \displaystyle{\frac{1}{2}\bigg(1-\frac{1}{7}\bigg).} \end{array}} |
| If we compare 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_1,s_2,s_3,} we notice a pattern. |
| 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=\frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg).} |
| Step 3: |
|---|
| Now, to calculate the sum of this series we need to 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 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{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{1}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\ &&\\ & = & \displaystyle{\frac{1}{2}.} \end{array}} |
| Since the partial sums converge, the series converges and the sum of the series is 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}.} |
| 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 \frac{8}{3}} |
| (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}} |