Difference between revisions of "009C Sample Final 2, Problem 3"
Jump to navigation
Jump to search
Kayla Murray (talk | contribs) |
Kayla Murray (talk | contribs) |
||
| Line 47: | Line 47: | ||
!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |For |
|- | |- | ||
| − | | | + | | <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1},</math> |
|- | |- | ||
| − | | | + | |we notice that this series is alternating. |
| + | |- | ||
| + | |Let <math style="vertical-align: -16px"> b_n=\frac{1}{n+1}.</math> | ||
| + | |- | ||
| + | |The sequence <math style="vertical-align: -5px">\{b_n\}</math> is decreasing since | ||
| + | |- | ||
| + | | <math>\frac{1}{n+2}<\frac{1}{n+1}</math> | ||
| + | |- | ||
| + | |for all <math style="vertical-align: -3px">n\ge 0.</math> | ||
|} | |} | ||
| Line 57: | Line 65: | ||
!Step 2: | !Step 2: | ||
|- | |- | ||
| − | | | + | |Also, |
| + | |- | ||
| + | | <math>\lim_{n\rightarrow \infty}b_n=\lim_{n\rightarrow \infty}\frac{1}{n+1}=0.</math> | ||
| + | |- | ||
| + | |Therefore, the series <math>\sum_{n=1}^\infty (-1)^n\frac{1}{n+1}</math> converges | ||
|- | |- | ||
| − | | | + | |by the Alternating Series Test. |
|} | |} | ||
| Line 68: | Line 80: | ||
| '''(a)''' | | '''(a)''' | ||
|- | |- | ||
| − | | '''(b)''' | + | | '''(b)''' converges |
|} | |} | ||
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | [[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 19:40, 4 March 2017
Determine if the following series converges or diverges. Please give your reason(s).
(a)
(b)
| Foundations: |
|---|
Solution:
(a)
| Step 1: |
|---|
| Step 2: |
|---|
(b)
| Step 1: |
|---|
| For |
| we notice that this series is alternating. |
| Let |
| The sequence is decreasing since |
| for all |
| Step 2: |
|---|
| Also, |
| Therefore, the series converges |
| by the Alternating Series Test. |
| Final Answer: |
|---|
| (a) |
| (b) converges |