Difference between revisions of "009C Sample Midterm 3, Problem 4"

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(Created page with "<span class="exam">Test the series for convergence or divergence. ::<span class="exam">(a) (6 points)      <math>{\displaystyle \sum_{n=1}^{\infty}}\,(-1...")
 
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{\displaystyle \lim_{n\rightarrow\infty}\left|(-1)^{n}\sin\left(\frac{\pi}{n}\right)\right|} & = & \left|\sin\left({\displaystyle \lim_{n\rightarrow\infty}\frac{\pi}{n}}\right)\right|\\
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& = & 0,
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\end{array} </math>
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|<br>as we can pass the limit through absolute value and sine because both are continuous functions. Since the series alternates for <math style="vertical-align: -15%">n\geq 2</math>, the series is convergent by the alternating series test.
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!(b): &nbsp;
 
!(b): &nbsp;
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|In this case, we find
 
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{\displaystyle \lim_{n\rightarrow\infty}\left|(-1)^{n}\cos\left(\frac{\pi}{n}\right)\right|} & = & \left|\cos\left({\displaystyle \lim_{n\rightarrow\infty}\frac{\pi}{n}}\right)\right|\\
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& = & \cos(0)\\
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& = & 1,
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\end{array}</math>
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|where we can again pass the limit through absolute value and cosine because both are continuous. Since the terms do not converge to zero absolutely, they do not converge to zero.  Thus, by the divergence test, the series is divergent.
 
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!Final Answer: &nbsp;
 
!Final Answer: &nbsp;
 
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|The series for (a) is convergent, while the series (b) is divergent.
 
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[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 15:21, 26 April 2015

Test the series for convergence or divergence.

(a) (6 points)     
(b) (6 points)     
Foundations:  
For , both sine and cosine of are strictly nonnegative. Thus, these series are alternating, and we can apply the
Alternating Series Test: If a series is
  • Alternating in sign, and
then the series is convergent.
Note that if the series does not converge to zero, we must claim it diverges by the

Divergence Test: If then the series/sum diverges.

In the case of an alternating series, such as the two listed for this problem, we can choose to show it does not converge to zero absolutely.

 Solution:

(a):  
Here, we have

as we can pass the limit through absolute value and sine because both are continuous functions. Since the series alternates for , the series is convergent by the alternating series test.
(b):  
In this case, we find
where we can again pass the limit through absolute value and cosine because both are continuous. Since the terms do not converge to zero absolutely, they do not converge to zero. Thus, by the divergence test, the series is divergent.
Final Answer:  
The series for (a) is convergent, while the series (b) is divergent.

Return to Sample Exam