Difference between revisions of "009C Sample Midterm 1, Problem 5"
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& = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\ | & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\ | ||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{|x-3|} | + | & = & \displaystyle{|x-3|.} |
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
| Line 142: | Line 142: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |The Ratio Test tells us this series is absolutely convergent if <math>|x-3|<1.</math> | + | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|x-3|<1.</math> |
|- | |- | ||
| − | |Hence, the Radius of Convergence of this series is <math>R=1.</math> | + | |Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math> |
|} | |} | ||
| Line 152: | Line 152: | ||
|Now, we need to determine the interval of convergence. | |Now, we need to determine the interval of convergence. | ||
|- | |- | ||
| − | |First, note that <math>|x-3|<1</math> corresponds to the interval <math>(2,4).</math> | + | |First, note that <math style="vertical-align: -5px">|x-3|<1</math> corresponds to the interval <math style="vertical-align: -4px">(2,4).</math> |
|- | |- | ||
|To obtain the interval of convergence, we need to test the endpoints of this interval | |To obtain the interval of convergence, we need to test the endpoints of this interval | ||
|- | |- | ||
| − | |for convergence since the Ratio Test is inconclusive when <math>R=1.</math> | + | |for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -2px">R=1.</math> |
|} | |} | ||
| Line 162: | Line 162: | ||
!Step 4: | !Step 4: | ||
|- | |- | ||
| − | |First, let <math>x=4.</math> | + | |First, let <math style="vertical-align: -1px">x=4.</math> |
|- | |- | ||
|Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math> | |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \frac{1}{2n+1}.</math> | ||
| Line 168: | Line 168: | ||
|This is an alternating series. | |This is an alternating series. | ||
|- | |- | ||
| − | |Let <math>b_n=\frac{1}{2n+1}.</math>. | + | |Let <math style="vertical-align: -16px">b_n=\frac{1}{2n+1}.</math>. |
|- | |- | ||
|The sequence <math>\{b_n\}</math> is decreasing since | |The sequence <math>\{b_n\}</math> is decreasing since | ||
| Line 174: | Line 174: | ||
| <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math> | | <math>\frac{1}{2(n+1)+1}<\frac{1}{2n+1}</math> | ||
|- | |- | ||
| − | |for all <math>n\ge 1.</math> | + | |for all <math style="vertical-align: -3px">n\ge 1.</math> |
|- | |- | ||
|Also, | |Also, | ||
| Line 182: | Line 182: | ||
|Therefore, this series converges by the Alternating Series Test | |Therefore, this series converges by the Alternating Series Test | ||
|- | |- | ||
| − | |and we include <math>x=4</math> in our interval. | + | |and we include <math style="vertical-align: -1px">x=4</math> in our interval. |
|} | |} | ||
| Line 188: | Line 188: | ||
!Step 5: | !Step 5: | ||
|- | |- | ||
| − | |Now, let <math>x=2.</math> | + | |Now, let <math style="vertical-align: -1px">x=2.</math> |
|- | |- | ||
|Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math> | |Then, the series becomes <math>\sum_{n=0}^\infty \frac{1}{2n+1}.</math> | ||
|- | |- | ||
| − | |First, we note that <math>\frac{1}{2n+1}>0</math> for all <math>n\ge 0.</math> | + | |First, we note that <math>\frac{1}{2n+1}>0</math> for all <math style="vertical-align: -3px">n\ge 0.</math> |
|- | |- | ||
|Thus, we can use the Limit Comparison Test. | |Thus, we can use the Limit Comparison Test. | ||
| Line 213: | Line 213: | ||
|<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> diverges by the | |<math>\sum_{n=0}^\infty \frac{1}{2n+1}</math> diverges by the | ||
|- | |- | ||
| − | |Limit Comparison Test. Therefore, we do not include <math>x=2</math> | + | |Limit Comparison Test. Therefore, we do not include <math style="vertical-align: -1px">x=2</math> |
|- | |- | ||
|in our interval. | |in our interval. | ||
| Line 221: | Line 221: | ||
!Step 6: | !Step 6: | ||
|- | |- | ||
| − | |The interval of convergence is <math>(2,4].</math> | + | |The interval of convergence is <math style="vertical-align: -4px">(2,4].</math> |
|} | |} | ||
| Line 228: | Line 228: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | '''(a)''' The radius of convergence is <math>R=1</math> and the interval of convergence is <math>(-1,1).</math> | + | | '''(a)''' The radius of convergence is <math style="vertical-align: -2px">R=1</math> and the interval of convergence is <math style="vertical-align: -4px">(-1,1).</math> |
|- | |- | ||
| − | | '''(b)''' The radius of convergence is <math>R=1</math> and the interval fo convergence is <math>(2,4].</math> | + | | '''(b)''' The radius of convergence is <math style="vertical-align: -2px">R=1</math> and the interval fo convergence is <math style="vertical-align: -4px">(2,4].</math> |
|} | |} | ||
[[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 14:57, 15 February 2017
Find the radius of convergence and interval of convergence of the series.
- 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 \sum_{n=0}^\infty \sqrt{n}x^n}
- 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 \sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}}
| Foundations: |
|---|
| 1. Ratio Test |
| 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 \sum a_n} be a series 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.} |
| Then, |
|
If 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 L<1,} the series is absolutely convergent. |
|
If 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 L>1,} the series is divergent. |
|
If 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 L=1,} the test is inconclusive. |
| 2. If a series absolutely converges, then it also converges. |
Solution:
(a)
| Step 1: |
|---|
| We first use the Ratio Test to determine the radius of convergence. |
| 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}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}x^{n+1}}{\sqrt{n}x^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{\sqrt{n+1}}{\sqrt{n}}x\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}|x|}\\ &&\\ & = & \displaystyle{|x|\lim_{n\rightarrow \infty} \sqrt{\frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{\lim_{n\rightarrow \infty} \frac{n+1}{n}}}\\ &&\\ & = & \displaystyle{|x|\sqrt{1}}\\ &&\\ &=& \displaystyle{|x|.} \end{array}} |
| Step 2: |
|---|
| The Ratio Test tells us this series is absolutely convergent if 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|<1.} |
| Hence, the Radius of Convergence of this 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 R=1.} |
| Step 3: |
|---|
| Now, we need to determine the interval of convergence. |
| First, note that 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|<1} corresponds to the interval 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 (-1,1).} |
| To obtain the interval of convergence, we need to test the endpoints of this interval |
| for convergence since the Ratio Test is inconclusive when 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 L=1.} |
| Step 4: |
|---|
| First, 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=1.} |
| Then, the series becomes 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=0}^\infty \sqrt{n}.} |
| We note that |
| 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} \sqrt{n}=\infty.} |
| Therefore, the series diverges by 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 term test. |
| Hence, we do not include 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=1} in the interval. |
| Step 5: |
|---|
| Now, 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=-1.} |
| Then, the series becomes 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=0}^\infty (-1)^n \sqrt{n}.} |
| Since 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} \sqrt{n}=\infty,} |
| 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 \lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.} |
| Therefore, the series diverges by 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 term test. |
| Hence, we do not include 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=-1 } in the interval. |
| Step 6: |
|---|
| The interval of convergence 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 (-1,1).} |
(b)
| Step 1: |
|---|
| We first use the Ratio Test to determine the radius of convergence. |
| 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}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x-3)^{n+1}}{2(n+1)+1}\frac{2n+1}{(-1)^n(x-3)^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x-3)\frac{2n+1}{2n+3}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} |x-3|\frac{2n+1}{2n+3}}\\ &&\\ & = & \displaystyle{|x-3|\lim_{n\rightarrow \infty} \frac{2n+1}{2n+3}}\\ &&\\ & = & \displaystyle{|x-3|.} \end{array}} |
| Step 2: |
|---|
| The Ratio Test tells us this series is absolutely convergent if 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-3|<1.} |
| Hence, the Radius of Convergence of this 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 R=1.} |
| Step 3: |
|---|
| Now, we need to determine the interval of convergence. |
| First, note that 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-3|<1} corresponds to the interval 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 (2,4).} |
| To obtain the interval of convergence, we need to test the endpoints of this interval |
| for convergence since the Ratio Test is inconclusive when 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 R=1.} |
| Step 4: |
|---|
| First, 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=4.} |
| Then, the series becomes 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=0}^\infty (-1)^n \frac{1}{2n+1}.} |
| This is an alternating series. |
| 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 b_n=\frac{1}{2n+1}.} . |
| The sequence 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_n\}} is decreasing since |
| 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(n+1)+1}<\frac{1}{2n+1}} |
| for all 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\ge 1.} |
| Also, |
| 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} b_n=\lim_{n\rightarrow \infty} \frac{1}{2n+1}=0.} |
| Therefore, this series converges by the Alternating Series Test |
| and we include 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=4} in our interval. |
| Step 5: |
|---|
| Now, 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=2.} |
| Then, the series becomes 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=0}^\infty \frac{1}{2n+1}.} |
| First, we note that 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}{2n+1}>0} for all 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\ge 0.} |
| Thus, we can use the Limit Comparison Test. |
| We compare this series with the 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 \frac{1}{n},} |
| which is the harmonic series and divergent. |
| Now, 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} \frac{\frac{1}{2n+1}}{\frac{1}{n}}} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n}{2n+1}}\\ &&\\ & = & \displaystyle{\frac{1}{2}.} \end{array}} |
| Since this limit is a finite number greater than zero, 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 \sum_{n=0}^\infty \frac{1}{2n+1}} diverges by the |
| Limit Comparison Test. Therefore, we do not include 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=2} |
| in our interval. |
| Step 6: |
|---|
| The interval of convergence 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 (2,4].} |
| Final Answer: |
|---|
| (a) The radius of convergence 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 R=1} and the interval of convergence 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 (-1,1).} |
| (b) The radius of convergence is and the interval fo convergence is |
![{\displaystyle (2,4].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45ad0009cd49d7732f72f4bfe26f00c0671fecc8)