Difference between revisions of "009C Sample Final 1, Problem 4"
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 1: | Line 1: | ||
<span class="exam"> Find the interval of convergence of the following series. | <span class="exam"> Find the interval of convergence of the following series. | ||
| − | + | ::<math>\sum_{n=0}^{\infty} (-1)^n \frac{(x+2)^n}{n^2}</math> | |
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
|- | |- | ||
| − | | | + | |'''1. 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> Then, |
| − | |||
|- | |- | ||
| | | | ||
| − | + | If <math style="vertical-align: -1px">L<1,</math> the series is absolutely convergent. | |
|- | |- | ||
| | | | ||
| − | + | If <math style="vertical-align: -1px">L>1,</math> the series is divergent. | |
|- | |- | ||
| | | | ||
| − | + | If <math style="vertical-align: -1px">L=1,</math> the test is inconclusive. | |
|- | |- | ||
| − | | | + | |'''2.''' After you find the radius of convergence, you need to check the endpoints of your interval |
| − | |||
|- | |- | ||
| | | | ||
| − | + | for convergence since the Ratio Test is inconclusive when <math style="vertical-align: -1px">L=1.</math> | |
|} | |} | ||
| + | |||
'''Solution:''' | '''Solution:''' | ||
| Line 35: | Line 34: | ||
|- | |- | ||
| | | | ||
| − | + | <math>\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+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\ | \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(-1)^{n+1}(x+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\ | ||
&&\\ | &&\\ | ||
| Line 64: | Line 63: | ||
|- | |- | ||
| | | | ||
| − | + | <math>\sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}.</math> | |
|- | |- | ||
|Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing. | |Since <math style="vertical-align: -5px">n^2<(n+1)^2,</math> we have <math>\frac{1}{(n+1)^2}<\frac{1}{n^2}.</math> Thus, <math>\frac{1}{n^2}</math> is decreasing. | ||
| Line 77: | Line 76: | ||
|- | |- | ||
| | | | ||
| − | + | <math>\begin{array}{rcl} | |
\displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ | \displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ | ||
&&\\ | &&\\ | ||
| Line 95: | Line 94: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | <math>[-3,-1]</math> | + | | <math>[-3,-1]</math> |
|} | |} | ||
[[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 15:59, 25 February 2017
Find the interval of convergence of the following series.
| Foundations: |
|---|
| 1. Ratio Test |
| Let 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. After you find the radius of convergence, you need to check the endpoints of your 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.} |
Solution:
| Step 1: |
|---|
| We proceed using the ratio test to find the interval of convergence. 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{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(-1)^{n+1}(x+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|(1)^2}\\ &&\\ & = & \displaystyle{|x+2|.}\\ \end{array}} |
| Step 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 |x+2|<1.} Hence, our interval 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 (-3,-1).} But, we still need to check the endpoints of this interval |
| to see if they are included in the interval of convergence. |
| Step 3: |
|---|
| First, 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=-1.} Then, our 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}{n^2}.} |
| 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 n^2<(n+1)^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 \frac{1}{(n+1)^2}<\frac{1}{n^2}.} Thus, 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}{n^2}} is decreasing. |
| So, 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}{n^2}} converges by the Alternating Series Test. |
| Step 4: |
|---|
| Now, 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=-3.} Then, our 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 \begin{array}{rcl} \displaystyle{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}.}\\ \end{array}} |
| This is a convergent series by the p-test. |
| Step 5: |
|---|
| Thus, the interval of convergence for 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 [-3,-1].} |
| Final Answer: |
|---|
| 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 [-3,-1]} |