Difference between revisions of "009C Sample Midterm 2, Problem 3"

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!Foundations:    
 
!Foundations:    
 
|-
 
|-
|  
+
|Alternating Series Test
 
|-
 
|-
|
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|Ratio Test
::
 
 
|-
 
|-
 
|
 
|
::
 
 
|}
 
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!Step 1:    
 
!Step 1:    
 
|-
 
|-
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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;
 
!Step 2: &nbsp;
 
|-
 
|-
|
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|We notice that the series is alternating.
 +
|-
 +
|Let <math> b_n=\frac{1}{\sqrt{n}}.</math>
 +
|-
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|The sequence <math>\{b_n\}</math> is decreasing since
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math>
 +
|-
 +
|for all <math>n\ge 1.</math>
 +
|-
 +
|Also,
 +
|-
 +
|&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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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
|'''(a)'''  
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; converges
 
|-
 
|-
 
|'''(b)'''  
 
|'''(b)'''  
 
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[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 12:05, 13 February 2017

Determine convergence or divergence:

a)
b)


Foundations:  
Alternating Series Test
Ratio Test

Solution:

(a)

Step 1:  
First, we have
       
Step 2:  
We notice that the series is alternating.
Let
The sequence is decreasing since
       
for all
Also,
       
Therefore, the series Failed to parse (syntax error): {\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n|}} converges by the Alternating Series Test.

(b)

Step 1:  
Step 2:  
Final Answer:  
    (a)     converges
(b)

Return to Sample Exam