Difference between revisions of "009C Sample Final 3, Problem 2"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 27: | Line 27: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |First, we take the absolute value of the terms in the original series. |
| + | |- | ||
| + | |Let <math style="vertical-align: -20px">a_n=\frac{(-1)^n}{\sqrt{n}}.</math> | ||
|- | |- | ||
| − | | | + | |Therefore, |
|- | |- | ||
| − | | | + | | <math>\begin{array}{rcl} |
| + | \displaystyle{\sum_{n=1}^\infty |a_n|} & = & \displaystyle{\sum_{n=1}^\infty \bigg|\frac{(-1)^n}{\sqrt{n}}\bigg|}\\ | ||
| + | &&\\ | ||
| + | & = & \displaystyle{\sum_{n=1}^\infty \frac{1}{\sqrt{n}}.} | ||
| + | \end{array}</math> | ||
|} | |} | ||
| Line 37: | Line 43: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |This series is a <math style="vertical-align: -5px">p</math>-series with <math style="vertical-align: -14px">p=\frac{1}{2}.</math> |
| + | |- | ||
| + | |Therefore, it diverges. | ||
| + | |- | ||
| + | |Hence, the series | ||
| + | |- | ||
| + | | <math>\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}</math> | ||
| + | |- | ||
| + | |is not absolutely convergent. | ||
|- | |- | ||
| | | | ||
| Line 66: | Line 80: | ||
!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | | '''(a)''' | + | | '''(a)''' not absolutely convergent |
|- | |- | ||
| '''(b)''' | | '''(b)''' | ||
|} | |} | ||
[[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_3|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 13:52, 5 March 2017
Consider the series
(a) Test if the series converges absolutely. Give reasons for your answer.
(b) Test if the series converges conditionally. Give reasons for your answer.
| Foundations: |
|---|
| 1. A series is absolutely convergent if |
| the series converges. |
| 2. A series is conditionally convergent if |
| the series diverges and the series converges. |
Solution:
(a)
| Step 1: |
|---|
| First, we take the absolute value of the terms in the original series. |
| Let |
| Therefore, |
| Step 2: |
|---|
| This series is a -series with |
| Therefore, it diverges. |
| Hence, the series |
| is not absolutely convergent. |
(b)
| Step 1: |
|---|
| Step 2: |
|---|
| Final Answer: |
|---|
| (a) not absolutely convergent |
| (b) |