Difference between revisions of "009C Sample Midterm 1, Problem 5"
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<span class="exam"> Find the radius of convergence and interval of convergence of the series. | <span class="exam"> Find the radius of convergence and interval of convergence of the series. | ||
| − | <span class="exam">(a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> | + | <span class="exam">(a) <math>\sum_{n=0}^\infty \sqrt{n}x^n</math> |
| − | <span class="exam">(b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math> | + | <span class="exam">(b) <math>\sum_{n=0}^\infty (-1)^n \frac{(x-3)^n}{2n+1}</math> |
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
| Line 10: | Line 10: | ||
|'''Ratio Test''' | |'''Ratio Test''' | ||
|- | |- | ||
| − | | Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> | + | | Let <math style="vertical-align: -7px">\sum a_n</math> be a series and <math>L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|.</math> |
|- | |- | ||
| Then, | | Then, | ||
|- | |- | ||
| | | | ||
| − | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent. | + | If <math style="vertical-align: -4px">L<1,</math> the series is absolutely convergent. |
|- | |- | ||
| | | | ||
| − | If <math style="vertical-align: -4px">L>1,</math> the series is divergent. | + | If <math style="vertical-align: -4px">L>1,</math> the series is divergent. |
|- | |- | ||
| | | | ||
| − | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive. | + | If <math style="vertical-align: -4px">L=1,</math> the test is inconclusive. |
|} | |} | ||
| Line 55: | Line 55: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|x|<1.</math> | + | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|x|<1.</math> |
|- | |- | ||
| − | |Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math> | + | |Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math> |
|} | |} | ||
| Line 65: | Line 65: | ||
|Now, we need to determine the interval of convergence. | |Now, we need to determine the interval of convergence. | ||
|- | |- | ||
| − | |First, note that <math style="vertical-align: -5px">|x|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-1,1).</math> | + | |First, note that <math style="vertical-align: -5px">|x|<1</math> corresponds to the interval <math style="vertical-align: -4px">(-1,1).</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 style="vertical-align: -2px">L=1.</math> | + | |for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -2px">L=1.</math> |
|} | |} | ||
| Line 75: | Line 75: | ||
!Step 4: | !Step 4: | ||
|- | |- | ||
| − | |First, let <math style="vertical-align: -1px">x=1.</math> | + | |First, let <math style="vertical-align: -1px">x=1.</math> |
|- | |- | ||
| − | |Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math> | + | |Then, the series becomes <math>\sum_{n=0}^\infty \sqrt{n}.</math> |
|- | |- | ||
|We note that | |We note that | ||
| Line 83: | Line 83: | ||
| <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math> | | <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty.</math> | ||
|- | |- | ||
| − | |Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test. | + | |Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test. |
|- | |- | ||
| − | |Hence, we do not include <math style="vertical-align: -1px">x=1</math> in the interval. | + | |Hence, we do not include <math style="vertical-align: -1px">x=1</math> in the interval. |
|} | |} | ||
| Line 91: | Line 91: | ||
!Step 5: | !Step 5: | ||
|- | |- | ||
| − | |Now, let <math style="vertical-align: -1px">x=-1.</math> | + | |Now, let <math style="vertical-align: -1px">x=-1.</math> |
|- | |- | ||
| − | |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math> | + | |Then, the series becomes <math>\sum_{n=0}^\infty (-1)^n \sqrt{n}.</math> |
|- | |- | ||
| − | |Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math> | + | |Since <math>\lim_{n\rightarrow \infty} \sqrt{n}=\infty,</math> |
|- | |- | ||
|we have | |we have | ||
| Line 101: | Line 101: | ||
| <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math> | | <math>\lim_{n\rightarrow \infty} (-1)^n\sqrt{n}=\text{DNE}.</math> | ||
|- | |- | ||
| − | |Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test. | + | |Therefore, the series diverges by the <math style="vertical-align: 0px">n</math>th term test. |
|- | |- | ||
| − | |Hence, we do not include <math style="vertical-align: -1px">x=-1 </math> in the interval. | + | |Hence, we do not include <math style="vertical-align: -1px">x=-1 </math> in the interval. |
|} | |} | ||
| Line 109: | Line 109: | ||
!Step 6: | !Step 6: | ||
|- | |- | ||
| − | |The interval of convergence is <math style="vertical-align: -4px">(-1,1).</math> | + | |The interval of convergence is <math style="vertical-align: -4px">(-1,1).</math> |
|} | |} | ||
| Line 137: | Line 137: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |The Ratio Test tells us this series is absolutely convergent if <math style="vertical-align: -5px">|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 style="vertical-align: -2px">R=1.</math> | + | |Hence, the Radius of Convergence of this series is <math style="vertical-align: -2px">R=1.</math> |
|} | |} | ||
| Line 147: | Line 147: | ||
|Now, we need to determine the interval of convergence. | |Now, we need to determine the interval of convergence. | ||
|- | |- | ||
| − | |First, note that <math style="vertical-align: -5px">|x-3|<1</math> corresponds to the interval <math style="vertical-align: -4px">(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 style="vertical-align: -2px">R=1.</math> | + | |for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -2px">R=1.</math> |
|} | |} | ||
| Line 157: | Line 157: | ||
!Step 4: | !Step 4: | ||
|- | |- | ||
| − | |First, let <math style="vertical-align: -1px">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> |
|- | |- | ||
|This is an alternating series. | |This is an alternating series. | ||
|- | |- | ||
| − | |Let <math style="vertical-align: -16px">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 |
|- | |- | ||
| <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 style="vertical-align: -3px">n\ge 1.</math> | + | |for all <math style="vertical-align: -3px">n\ge 1.</math> |
|- | |- | ||
|Also, | |Also, | ||
| Line 177: | Line 177: | ||
|Therefore, this series converges by the Alternating Series Test | |Therefore, this series converges by the Alternating Series Test | ||
|- | |- | ||
| − | |and we include <math style="vertical-align: -1px">x=4</math> in our interval. | + | |and we include <math style="vertical-align: -1px">x=4</math> in our interval. |
|} | |} | ||
| Line 183: | Line 183: | ||
!Step 5: | !Step 5: | ||
|- | |- | ||
| − | |Now, let <math style="vertical-align: -1px">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 style="vertical-align: -3px">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. | ||
|- | |- | ||
| − | |We compare this series with the series <math>\sum_{n=1}^\infty \frac{1}{n},</math> | + | |We compare this series with the series <math>\sum_{n=1}^\infty \frac{1}{n},</math> |
|- | |- | ||
|which is the harmonic series and divergent. | |which is the harmonic series and divergent. | ||
| Line 206: | Line 206: | ||
|Since this limit is a finite number greater than zero, we have | |Since this limit is a finite number greater than zero, we have | ||
|- | |- | ||
| − | |<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 style="vertical-align: -1px">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 216: | Line 216: | ||
!Step 6: | !Step 6: | ||
|- | |- | ||
| − | |The interval of convergence is <math style="vertical-align: -4px">(2,4].</math> | + | |The interval of convergence is <math style="vertical-align: -4px">(2,4].</math> |
|} | |} | ||
| Line 223: | Line 223: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | '''(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> | + | | '''(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 style="vertical-align: -2px">R=1</math> and the interval | + | | '''(b)''' The radius of convergence is <math style="vertical-align: -2px">R=1</math> and the interval of 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 17:59, 26 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: |
|---|
| 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. |
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 of convergence is |
![{\displaystyle (2,4].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/45ad0009cd49d7732f72f4bfe26f00c0671fecc8)