Does the Alternating Series Converge?

Click For Summary

Discussion Overview

The discussion revolves around the convergence properties of the alternating series given by the sum \(\Sigma(-1)^{k}\frac{(k+4)}{(k^{2}+k)}\). Participants explore whether the series converges, diverges, or conditionally converges, with a focus on applying the Alternating Series test and testing for absolute convergence.

Discussion Character

  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant suggests testing for absolute convergence by comparing the series to \(\frac{1}{k}\), concluding that it diverges.
  • Another participant questions whether the sequence \(\frac{(k+4)}{(k^{2}+k)}\) converges to 0 as \(k\) approaches infinity, proposing to simplify the expression by dividing by \(k^2\).
  • A participant responds affirmatively, indicating that the limit approaches 4, but does not clarify the implications for convergence.
  • Another participant emphasizes the need to show that \(a_{n+1} < a_n\) for conditional convergence, referencing a previous suggestion for evaluating the limit of the sequence.
  • One participant corrects a misunderstanding regarding the limit, indicating that the terms should approach \(\frac{1}{k^2}\) rather than 4.

Areas of Agreement / Disagreement

Participants generally agree on the need to test for conditional convergence after establishing that the series does not absolutely converge. However, there is disagreement on the evaluation of limits and the implications for convergence, indicating that the discussion remains unresolved.

Contextual Notes

Limitations include potential misunderstandings in the evaluation of limits and the conditions required for applying the Alternating Series test. The discussion reflects varying interpretations of the convergence criteria.

Heat
Messages
272
Reaction score
0
I need help identifying if it converges or diverges or conditionally converges.

\Sigma(-1)^{k}\frac{(k+4)}{(k^{2}+k)}


First I want to test for absolute convergence, and comparing this limit to 1/k I get that it diverges. Since it diverges, I need to test it now using the Alternating Series test, in which the limit is 0 and it converges. So my question is, does it conditionally converge?
 
Physics news on Phys.org
So your question is just whether the sequence (k+4)/(k2+ k) converges to 0 as k goes to infinity? Divide both numerator and denominator by k2. Now what happens as k goes to infinity?
 
yes from 0 to infinity.

it would be (1/k + 4/k)(1+1/k)...4/1 = 4...so?
 
Ok, you've established that it does not absolutely converge so you must test for conditional convergence, but your test is not comprehensive. Designate a_sub_n as (k+4)/(k^2 + k) and taking HallsofIvy's suggestion, we have lim(n -> +inf) of a_sub_n = lim(n -> +inf) of (1/k + 4/k^2)/(1 + 1/k), which tends to 0 because all the terms with k in the denominator go to 0. Now you have to show that a_sub_(n+1) < a_sub_n, and if this is true, then you can say that the series conditionally converges.
 
Heat said:
yes from 0 to infinity.

it would be (1/k + 4/k)(1+1/k)...4/1 = 4...so?

This is not correct. Note, that the 4 in the numerator comes as 4/k ...in fact it should be 1/k^2.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K