Series Comparison Test, help?

In summary, the statements regarding the convergence or divergence of different series using the Comparison Test were discussed. The arguments for (1), (4), and (5) were determined to be incorrect, while the arguments for (2), (3), and (6) were found to be valid.
  • #1
ani9890
11
0
Series Comparison Test, URGENT help?

Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter I (for "incorrect") if any part of the argument is flawed.

1. For all n>2, ∑ 1/ (n^2−1) < 1/n^2 so converges
2. For all n> 1, ∑ arctan(n) / n^3 < pi / 2n^3 so converges
3. For all n>1, ∑ ln(n)/n^2 < 1/n^1.5 so converges
4. For all n>1, ∑ 1/nln(n) < 2/n so diverges
5. For all n>2, ∑ n/(n^3 - 8) < 2/n^2 so converges
6. For all n>2, ∑ ln(n)/n > 1/n so diverges

I believe (1) is incorrect.
(2), (3), and (6) I believe are correct.
But I can't figure out 4 and 5 ? I think 4 is incorrect but I'm not sure, and for 5 shouldn't it be compared to 1/n^2 ?

Please help!
 
Last edited:
Physics news on Phys.org
  • #2


What's your reasoning behind the numbers you have answers for?
 
  • #3


for (1) 1/n^2 is actually greater than 1/ (n^2−1), for (4) if it diverges it should actually be asubn > bsubn not the other way around, and (5) the bsubn should be n/n^3 which becomes 1/n^2

for the (2),(3) 1/ n^p where p>1 so it converges which is correct, and for (6) n^p, here p=1 so it diverges which is correct.
 
  • #4


Your reasoning for 1 is correct (the same reasoning applies to 5; that one has the inequality in the wrong direction).
2, 3, and 4 are correct too.
 
  • #5


thank you for your help,
so I just wanted to make sure, for my answers:

1. incorrect
2. correct
3. correct
4. incorrect
5. incorrect
6. correct

is this okay?
 
  • #6


Looks good :smile:
 

Related to Series Comparison Test, help?

1. What is the Series Comparison Test and how does it work?

The Series Comparison Test is a method used to determine the convergence or divergence of an infinite series. It works by comparing the given series with a known series that has already been proven to converge or diverge.

2. When should I use the Series Comparison Test?

The Series Comparison Test should be used when the terms of the given series are positive and decreasing, and when the known series used for comparison is simpler to evaluate than the given series.

3. What is the main difference between the Limit Comparison Test and the Series Comparison Test?

The main difference between the Limit Comparison Test and the Series Comparison Test is that the Limit Comparison Test compares the ratios of the terms of two series, while the Series Comparison Test compares the actual terms of two series.

4. Can the Series Comparison Test be used to prove a series is absolutely convergent?

Yes, the Series Comparison Test can be used to prove absolute convergence by comparing the given series with a known absolutely convergent series. If the known series converges absolutely, then the given series must also converge absolutely.

5. Are there any common mistakes when using the Series Comparison Test?

Yes, some common mistakes when using the Series Comparison Test include not checking the necessary conditions for using the test, such as positive and decreasing terms, and using an incorrect known series for comparison.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
242
  • Calculus and Beyond Homework Help
Replies
1
Views
311
  • Calculus and Beyond Homework Help
Replies
2
Views
730
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
29
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
445
Back
Top