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) &nbsp;<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) &nbsp;<math>\sum_{n=1}^\infty (-2)^n \frac{n!}{n^n} </math>
  
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<hr>
!Foundations: &nbsp;
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[[009C Sample Midterm 2, Problem 3 Solution|'''<u>Solution</u>''']]
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|Alternating Series Test
 
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|Ratio Test
 
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'''Solution:'''
 
  
'''(a)'''
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[[009C Sample Midterm 2, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']]
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!Step 1: &nbsp;
 
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|First, we have
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\sum_{n=1}^\infty (-1)^n \sqrt{\frac{1}{n}}=\sum_{n=1}^\infty (-1)^n \frac{1}{\sqrt{n}}.</math>  
 
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!Step 2: &nbsp;
 
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|We notice that the series is alternating.
 
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|Let <math> b_n=\frac{1}{\sqrt{n}}.</math>
 
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|The sequence <math>\{b_n\}</math> is decreasing since
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 
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|for all <math>n\ge 1.</math>
 
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|Also,
 
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}=0.</math>
 
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|Therefore, the series <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> converges by the Alternating Series Test.
 
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'''(b)'''
 
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!Step 1: &nbsp;
 
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!Step 2: &nbsp;
 
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!Final Answer: &nbsp;
 
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|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges
 
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|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; converges
 
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[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Latest revision as of 12:35, 12 November 2017

Determine convergence or divergence:

(a)  

(b)  


Solution


Detailed Solution


Return to Sample Exam