What Happens When Your Comparison Test Fails or Is Negative?

  • Thread starter Thread starter Sabricd
  • Start date Start date
  • Tags Tags
    Series
Click For Summary

Homework Help Overview

The discussion revolves around the comparison test in the context of series convergence, particularly focusing on scenarios where the test fails or yields negative results. Participants are exploring the implications of such outcomes and their relationship to Maclaurin series.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster questions the consequences of a failed comparison test and whether negative or zero results are possible. Other participants inquire about specific series and offer clarifications on simplifications related to factorials and series terms.

Discussion Status

Participants are actively engaging with the original poster's concerns, providing insights into the nature of series and factorials. There is a focus on understanding the simplification process, though no consensus or resolution has been reached regarding the comparison test's implications.

Contextual Notes

Some participants mention a lack of experience with factorials and series, indicating that the discussion is occurring within a self-study context of calculus, specifically Calculus II.

Sabricd
Messages
27
Reaction score
0
what happens if your comparison test fails and you limit comparison test is negative or zero? Does that ever happen?
 
Physics news on Phys.org
Hi Sabricd! :wink:

If you choose the wrong comparison, that's bound to happen. :confused:

Is there a particular series that's worrying you? :smile:​
 
Hello!

Yes! This is probably silly but I am not sure how to simplify...

=1 + 2/(1!)*x + 6/(2!)*x^2 + 24/(3!)*x^3 + 120/(4!)*x^4 ...

I am trying to solve a Maclaurin series but I don't have that much experience with factorials. I would really REALLY appreciate if you could give me some insight!

Thanks!
 
Your series is actually

1+2x+3x^2+4x^3+...+(n+1)x^n+...

thats a little simpler, isn't it?
 
Yes!
Sorry I'm still kind of lost. I'm teaching myself Cal2..but I cannot understand how that is derived. Would you mind explaining to me how you simplified it?

Thanks!
 
Well, the numerators of your series are 1,2,6,24,120,... The general term is certainly (n+1)!
So your series becomes \frac{(n+1)!}{n!}x^n. Working this out gives you the answer.
 
Ok thanks! :)
 

Similar threads

Using comparison tests and limit comparison test
  • · Replies 1 ·
Replies
1
Views
2K
Determining if series converges or diverges
  • · Replies 5 ·
Replies
5
Views
2K
Proving convergence and divergence of series
  • · Replies 3 ·
Replies
3
Views
2K
Determining if a series converges
  • · Replies 4 ·
Replies
4
Views
2K
Proving the convergence of series
  • · Replies 14 ·
Replies
14
Views
2K
How to show that these sequences are summable?
  • · Replies 1 ·
Replies
1
Views
2K
Is Using the Comparison Test to Prove Divergence Valid for This Series?
  • · Replies 29 ·
Replies
29
Views
3K
Using the limit comparison test to prove conv or div
  • · Replies 10 ·
Replies
10
Views
3K
Comparison test of infinite series
  • · Replies 6 ·
Replies
6
Views
2K
Convergence or Divergence of Infinite Series: Methods and Examples
  • · Replies 4 ·
Replies
4
Views
2K