Difference between revisions of "009C Sample Midterm 2, Problem 3"
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!Foundations: | !Foundations: | ||
|- | |- | ||
− | | | + | |Alternating Series Test |
|- | |- | ||
− | | | + | |Ratio Test |
− | |||
|- | |- | ||
| | | | ||
− | |||
|} | |} | ||
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!Step 1: | !Step 1: | ||
|- | |- | ||
− | | | + | |First, we have |
|- | |- | ||
− | | | + | | <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: | !Step 2: | ||
|- | |- | ||
− | | | + | |We notice that the series is alternating. |
+ | |- | ||
+ | |Let <math> b_n=\frac{1}{\sqrt{n}}.</math> | ||
+ | |- | ||
+ | |The sequence <math>\{b_n\}</math> is decreasing since | ||
+ | |- | ||
+ | | <math>\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}</math> | ||
+ | |- | ||
+ | |for all <math>n\ge 1.</math> | ||
+ | |- | ||
+ | |Also, | ||
+ | |- | ||
+ | | <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. |
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | |'''(a)''' | + | | '''(a)''' converges |
|- | |- | ||
|'''(b)''' | |'''(b)''' | ||
|} | |} | ||
[[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: |
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Alternating Series Test |
Ratio Test |
Solution:
(a)
Step 1: |
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First, we have |
Step 2: |
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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: |
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Step 2: |
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Final Answer: |
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(a) converges |
(b) |