Difference between revisions of "009C Sample Final 2, Problem 3"
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− | <span class="exam">Determine if the following series converges or diverges. Please give your reason(s). | + | <span class="exam"> Determine if the following series converges or diverges. Please give your reason(s). |
− | <span class="exam">(a) <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math> | + | <span class="exam">(a) <math>\sum_{n=1}^{\infty} \frac{n!}{(2n)!}</math> |
− | <span class="exam">(b) <math>\sum_{n=1}^{\infty} (-1)^n \frac{1}{n+1}</math> | + | <span class="exam">(b) <math>\sum_{n=1}^{\infty} (-1)^n\frac{1}{n+1}</math> |
{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
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| | | | ||
<math>\begin{array}{rcl} | <math>\begin{array}{rcl} | ||
− | \displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!} | + | \displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)!}{(2(n+1))!} \frac{(2n)!}{n!}\bigg|}\\ |
&&\\ | &&\\ | ||
− | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} | + | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg| \frac{(n+1)n!}{(2n+2)(2n+1)(2n)!} \frac{(2n)!}{n!}\bigg|}\\ |
&&\\ | &&\\ | ||
& = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\ | & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{n+1}{(2n+2)(2n+1)}}\\ | ||
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|- | |- | ||
|Let <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math> | |Let <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math> | ||
+ | |- | ||
+ | |First, we have | ||
+ | |- | ||
+ | | <math>\frac{1}{n+1}\ge 0</math> | ||
+ | |- | ||
+ | |for all <math style="vertical-align: -3px">n\ge 1.</math> | ||
|- | |- | ||
|The sequence <math style="vertical-align: -5px">\{b_n\}</math> is decreasing since | |The sequence <math style="vertical-align: -5px">\{b_n\}</math> is decreasing since | ||
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| <math>\frac{1}{n+2}<\frac{1}{n+1}</math> | | <math>\frac{1}{n+2}<\frac{1}{n+1}</math> | ||
|- | |- | ||
− | |for all <math style="vertical-align: -3px">n\ge | + | |for all <math style="vertical-align: -3px">n\ge 1.</math> |
|} | |} | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
− | | '''(a)''' converges | + | | '''(a)''' converges (by the Ratio Test) |
|- | |- | ||
− | | '''(b)''' converges | + | | '''(b)''' converges (by the Alternating Series Test) |
|} | |} | ||
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] |
Revision as of 11:16, 17 March 2017
Determine if the following series converges or diverges. Please give your reason(s).
(a)
(b)
Foundations: |
---|
1. Ratio Test |
Let be a series and |
Then, |
If the series is absolutely convergent. |
If the series is divergent. |
If the test is inconclusive. |
2. If a series absolutely converges, then it also converges. |
3. Alternating Series Test |
Let be a positive, decreasing sequence where |
Then, and |
converge. |
Solution:
(a)
Step 1: |
---|
We begin by using the Ratio Test. |
We have |
|
Step 2: |
---|
Since |
the series is absolutely convergent by the Ratio Test. |
Therefore, the series converges. |
(b)
Step 1: |
---|
For |
we notice that this series is alternating. |
Let |
First, we have |
for all |
The sequence is decreasing since |
for all |
Step 2: |
---|
Also, |
Therefore, the series converges |
by the Alternating Series Test. |
Final Answer: |
---|
(a) converges (by the Ratio Test) |
(b) converges (by the Alternating Series Test) |