Converge absolutely or conditionally, or diverge?

  • Thread starter Thread starter rcmango
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves determining the convergence behavior of a series, specifically whether it converges absolutely, conditionally, or diverges. The series appears to be alternating and involves terms that may not approach zero.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the series, noting its alternating characteristics and questioning whether the terms approach zero. There is mention of using the nth term test for divergence and confusion regarding the behavior of terms involving roots.

Discussion Status

The discussion is active, with participants exploring various tests for convergence and divergence. Some have provided guidance on evaluating limits, while others are questioning the implications of their findings regarding the series' behavior.

Contextual Notes

There is uncertainty regarding the application of convergence tests, particularly in relation to the terms involving roots and their limits. Participants are also considering the relationship between the series in question and known divergent series like the harmonic series.

rcmango
Messages
232
Reaction score
0

Homework Statement



heres the problem: http://img145.imageshack.us/img145/161/55180818ir8.png

does this series converge absolutely, conditionally, or does it diverge?

Homework Equations





The Attempt at a Solution



whats series to test it with?
at first glance it looks like an alternating series. reminds me of a failing p series in the denominator.
 
Last edited by a moderator:
Physics news on Phys.org
The very, very simplest of convergence tests: if a series converges the individual terms must go to 0. Does that happen here? What is
lim_{n\rightarrow \infty} \frac{(-1)^n}{10^{1/n}}?
 
Okay, it appears to be alternating, however it is not decreasing because of the denominator, it is increasing!

thus diverging by the alternating series test, correct?
 
What he's using there is the nth term test for divergence. It should have been the first test that you learned, no?
 
yes, an easy test, how do i take the nth term test to a variable with a n root? the 10^1/n is confusing me. please help.
 
No, what I said was "If the individual terms do not go to 0, then the series must diverge"!

What is the limit of \frac{(-1)^n}{10^{(1/n)}}? Is it 0?

If you have trouble with that, what about 10^{1/n} itself? Calculate a few values, say n= 10 and higher.
 
i definitely can see that this does not go to 0.

However, how do I prove this on paper?

It fails convergence obviously.

Can i use direct comparison with harmonic series, and say that the harmonic series is smaller or equal to this series, thus showing if the harmonic series diverges, then the original series diverges.
 
rcmango said:
However, how do I prove this on paper?

Just take the limit of the summand as n goes to infinity. You don't need to compare the series with anything.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
2
Views
1K