Difference between revisions of "009C Sample Midterm 2, Problem 3"
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<span class="exam">Determine convergence or divergence: | <span class="exam">Determine convergence or divergence: | ||
| − | <span class="exam">(a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math> | + | <span class="exam">(a) <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}</math> |
| − | <span class="exam">(b) <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math> | + | <span class="exam">(b) <math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math> |
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|'''1.''' '''Alternating Series Test''' | |'''1.''' '''Alternating Series Test''' | ||
|- | |- | ||
| − | | Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math> | + | | Let <math>\{a_n\}</math> be a positive, decreasing sequence where <math style="vertical-align: -11px">\lim_{n\rightarrow \infty} a_n=0.</math> |
|- | |- | ||
| − | | Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math> | + | | Then, <math>\sum_{n=1}^\infty (-1)^na_n</math> and <math>\sum_{n=1}^\infty (-1)^{n+1}a_n</math> |
|- | |- | ||
| converge. | | converge. | ||
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|'''2.''' '''Ratio Test''' | |'''2.''' '''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. |
|- | |- | ||
|'''3.''' If a series absolutely converges, then it also converges. | |'''3.''' If a series absolutely converges, then it also converges. | ||
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|We notice that the series is alternating. | |We notice that the series is alternating. | ||
|- | |- | ||
| − | |Let <math> b_n=\frac{1}{\sqrt{n}}.</math> | + | |Let <math> b_n=\frac{1}{\sqrt{n}}.</math> |
|- | |- | ||
| − | |The sequence <math>\{b_n\}</math> is decreasing since | + | |The sequence <math>\{b_n\}</math> is decreasing since |
|- | |- | ||
| <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math> | | <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</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, | ||
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| <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math> | | <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math> | ||
|- | |- | ||
| − | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges by the Alternating Series Test. | + | |Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges by the Alternating Series Test. |
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |Now, we need to calculate <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math> | + | |Now, we need to calculate <math style="vertical-align: -15px">\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math> |
|- | |- | ||
| − | |Let <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math> | + | |Let <math style="vertical-align: -15px">y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.</math> |
|- | |- | ||
|Then, taking the natural log of both sides, we get | |Then, taking the natural log of both sides, we get | ||
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|since we can interchange limits and continuous functions. | |since we can interchange limits and continuous functions. | ||
|- | |- | ||
| − | |Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math> | + | |Now, this limit has the form <math style="vertical-align: -13px">\frac{0}{0}.</math> |
|- | |- | ||
|Hence, we can use L'Hopital's Rule to calculate this limit. | |Hence, we can use L'Hopital's Rule to calculate this limit. | ||
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!Step 4: | !Step 4: | ||
|- | |- | ||
| − | |Since <math>\ln y=-1,</math> we know | + | |Since <math>\ln y=-1,</math> we know |
|- | |- | ||
| <math>y=e^{-1}.</math> | | <math>y=e^{-1}.</math> | ||
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| <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math> | | <math>\lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.</math> | ||
|- | |- | ||
| − | |Since <math style="vertical-align: -13px">\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test. | + | |Since <math style="vertical-align: -13px">\frac{2}{e}<1,</math> the series is absolutely convergent by the Ratio Test. |
|- | |- | ||
|Therefore, the series converges. | |Therefore, the series converges. | ||
Revision as of 18:08, 26 February 2017
Determine convergence or divergence:
(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=1}^\infty (-1)^n \sqrt{\frac{1}{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=1}^\infty (-2)^n \frac{n!}{n^n} }
| Foundations: |
|---|
| 1. Alternating Series 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 \{a_n\}} be a positive, decreasing sequence where 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} a_n=0.} |
| Then, 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 (-1)^na_n} 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 \sum_{n=1}^\infty (-1)^{n+1}a_n} |
| converge. |
| 2. 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. |
| 3. If a series absolutely converges, then it also converges. |
Solution:
(a)
| Step 1: |
|---|
| 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 \sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.} |
| Step 2: |
|---|
| We notice that the series is alternating. |
| 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}{\sqrt{n}}.} |
| 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}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}} |
| 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}{\sqrt{n}}=0.} |
| Therefore, 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}{\sqrt{n}}} converges by the Alternating Series Test. |
(b)
| Step 1: |
|---|
| We begin by using the Ratio Test. |
| 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{(-2)^{n+1} (n+1)!}{(n+1)^{n+1}} \frac{n^n}{(-2)^n n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| (-2)(n+1) \frac{n^n}{(n+1)^{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} 2\frac{n^n}{(n+1)^n}}\\ &&\\ & = & \displaystyle{2\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.} \end{array}} |
| Step 2: |
|---|
| Now, 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} \bigg(\frac{n}{n+1}\bigg)^n.} |
| 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 y=\lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n.} |
| Then, taking the natural log of both sides, 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{\ln y } & = & \displaystyle{\ln \bigg( \lim_{n\rightarrow \infty} \bigg(\frac{n}{n+1}\bigg)^n \bigg)}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \ln \bigg(\frac{n}{n+1}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} n \ln \bigg(\frac{n}{n+1}\bigg) }\\ &&\\ & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}} \end{array}} |
| since we can interchange limits and continuous functions. |
| Now, this limit has the form 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{0}{0}.} |
| Hence, we can use L'Hopital's Rule to calculate this limit. |
| Step 3: |
|---|
| 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{\ln y } & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{\ln \bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{\ln \bigg(\frac{x}{x+1}\bigg)}{\frac{1}{x}}}\\ &&\\ & \overset{L'H}{=} & \displaystyle{\lim_{x\rightarrow \infty} \frac{\frac{1}{\bigg(\frac{x}{x+1}\bigg)}\frac{1}{(x+1)^2}}{\big(-\frac{1}{x^2}\big)}}\\ &&\\ & = & \displaystyle{\lim_{x\rightarrow \infty} \frac{-x}{x+1}}\\ &&\\ & = & \displaystyle{-1.} \end{array}} |
| Step 4: |
|---|
| 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 \ln y=-1,} we know |
| 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 y=e^{-1}.} |
| 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 \lim_{n\rightarrow \infty} \bigg|\frac{a_{n+1}}{a_n}\bigg|=2e^{-1}=\frac{2}{e}.} |
| 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{2}{e}<1,} the series is absolutely convergent by the Ratio Test. |
| Therefore, the series converges. |
| Final Answer: |
|---|
| (a) converges |
| (b) converges |