Converge conditionally or converge absolutely

  • Context: MHB 
  • Thread starter Thread starter shamieh
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the convergence properties of the alternating series $$\sum^{\infty}_{n= 1} \frac{(-1)^n}{n^2}$$. Participants explore whether this series converges conditionally, diverges, or converges absolutely, referencing various convergence tests and properties of related series.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants note that the series $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$ and express uncertainty about the behavior of the alternating series.
  • One participant suggests using L'Hôpital's rule to analyze the limit of $$\frac{1}{n}$$ and proposes applying the Alternating Series Test (AST) to conclude convergence.
  • Another participant points out that since the absolute series $$\sum \left| \frac{ (-1)^n }{n^2} \right| = \sum \frac{1}{n^2}$$ is convergent, the series is absolutely convergent.
  • One participant mentions the possibility of using the AST due to the nature of $$\frac{1}{n^2}$$ approaching zero as $$n$$ approaches infinity, and also references the p-series test indicating that $$p > 1$$ leads to convergence.
  • Another participant confirms that the Last Absolute Series Test (LAST) can also be applied to determine convergence.

Areas of Agreement / Disagreement

Participants express differing views on whether the series converges conditionally or absolutely, with some asserting absolute convergence while others focus on the need to test for absolute convergence first. The discussion remains unresolved regarding the final classification of the series.

Contextual Notes

Some participants highlight the need to clarify the distinction between absolute and conditional convergence, indicating that the approach to determining convergence may depend on the results of initial tests.

shamieh
Messages
538
Reaction score
0
I'm confused. I know that $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$ but I don't know what it does when it is alternating...

I'm given this $$\sum^{\infty}_{n= 1} \frac{(-1)^n}{n^2} $$ and told to find out if it conditionally converges, diverges or converges absolutely.Can I use lopitals to get 1/n and then use the AST and say 0 <= 1/n+1 <= 1/n and lim n-> infinity = 0 therefore by AST the series converges?
 
Last edited:
Physics news on Phys.org
shamieh said:
I'm confused. I know that $$\sum^{\infty}_{n = 1} \frac{1}{n^2}$$ converges to $$\frac{\pi^2}{6}$$ but I don't know what it does when it is alternating...

I'm given this $$\sum^{\infty}_{n= 1} \frac{(-1)^n}{n^2} $$ and told to find out if it conditionally converges, diverges or converges absolutely.Can I use lopitals to get 1/n and then use the AST and say 0 <= 1/n+1 <= 1/n and lim n-> infinity = 0 therefore by AST the series converges?

Since you already know that $\displaystyle \begin{align*} \sum \left| \frac{ (-1)^n }{n^2} \right| = \sum \frac{1}{n^2} \end{align*}$ is convergent, the series is obviously absolutely convergent...
 
Wow I'm making these way harder than they need to be. I need to get better at inspection. But couldn't you also use the AST test since you really just have a $$\frac{1}{n^2}$$ which is a number over infinity, which = 0 ? Also couldn't you just look at $$1/n^2$$ using the p series test as well to see that $$p > 1$$ so converges?
 
Yes LAST can be used to determine the convergence of this series as well.

HOWEVER, your question is asking you to determine if the series is ABSOLUTELY convergent or CONDITIONALLY convergent. That means you need to test for absolute convergence first. You would go for LAST if you couldn't show the series is absolutely convergent.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K