Convergence or Divergence of Infinite Series: Methods and Examples

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of the infinite series Ʃ(1/(n*ln(n)^2 - n)) from n = 1 to infinity. Participants explore various convergence tests including the Comparison Test, Integral Test, and Limit Comparison Test, while clarifying the nature of the series based on its denominator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the applicability of different convergence tests, express confusion about the denominator's form, and question the implications of negative terms in the series. Some suggest focusing on the tail end of the series for convergence analysis.

Discussion Status

The discussion is active with participants seeking clarification on the methods and definitions related to convergence tests. Some guidance has been offered regarding the use of the Integral Test and the importance of analyzing the series from a certain point onward. There is no explicit consensus yet on the best approach.

Contextual Notes

There is a noted concern about the series being negative for the first few terms, which raises questions about the validity of certain tests. Participants are also considering the implications of the series' behavior as n approaches infinity.

physicsernaw
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Homework Statement



Determine convergence or divergence using any method covered so far*:

Ʃ(1/(n*ln(n)^2 - n)) from n = 1 to infinity*The methods are the following:

- Dichotomy for positive series (if the partial sums are bounded above and the series is positive, the series converges)

- Integral Test

- Convergence of p-Series

- Comparison test

- Limit Comparison Test

Homework Equations



n/a

The Attempt at a Solution



The series is negative for n = 1 and 2, so I am left with Comparison test but I am having trouble determining what sequence to compare to. I am not "seeing" it.
 
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physicsernaw said:

Homework Statement



Determine convergence or divergence using any method covered so far*:

Ʃ(1/(n*ln(n)^2 - n)) from n = 1 to infinity

Is that denominator ##n\ln^2(n) -n## or ##n\ln (n^2)-n##?
*The methods are the following:

- Dichotomy for positive series (if the partial sums are bounded above and the series is positive, the series converges)

- Integral Test

- Convergence of p-Series

- Comparison test

- Limit Comparison Test

Homework Equations



n/a

The Attempt at a Solution



The series is negative for n = 1 and 2, so I am left with Comparison test but I am having trouble determining what sequence to compare to. I am not "seeing" it.

It doesn't matter if the first few terms of the series are negative. It is the tail end of the series that determines convergence or divergence. So don't rule out the other methods.
 
The denominator is the first one you posted, n*(ln(n))^2 - n.

In my book for the Limit Comparison Test it states "Let a[n] and b[n] be positive sequences." I assumed that "positive sequences" meant positive for all n [1, infinity). So it is really saying that if the sequence converges to a positive value or diverges "positively" than one can use the Limit Comp test on these sequences? Or am I still off base
 
physicsernaw said:
The denominator is the first one you posted, n*(ln(n))^2 - n.

In my book for the Limit Comparison Test it states "Let a[n] and b[n] be positive sequences." I assumed that "positive sequences" meant positive for all n [1, infinity). So it is really saying that if the sequence converges to a positive value or diverges "positively" than one can use the Limit Comp test on these sequences? Or am I still off base

If ##\sum_{n=3}^\infty a_n## converges or diverges, adding ##a_1+a_2## to it won't change the convergence or divergence. So analyze the sum from 3. I would suggest a combination of comparison and integral tests.
 
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Ahh I see thanks for clarifying. I will reattempt the problem.

EDIT: Got it by simply using integral test from n=3 to infinity. The integral converges so the series must as well. Thanks LCKurtz.
 
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