Solve Sequence Problem: (2n-1)/(3n^2 +1), n= 1,2,3,...

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Discussion Overview

The discussion revolves around the convergence or divergence of the sequence defined by the expression (2n-1)/(3n^2 +1) for n = 1, 2, 3, ... Participants explore methods to determine the limit of the sequence as n approaches infinity.

Discussion Character

  • Exploratory, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant suggests that the sequence converges to 0 and seeks confirmation of this claim.
  • Another participant proposes taking the limit as n approaches infinity and discusses the properties of the sequence being positive and decreasing for n > 2, arguing that this implies convergence to 0.
  • A different participant mentions using L'Hopital's rule to find the limit, initially confusing the context with the "nth term test" which is applicable to series.
  • Another participant critiques the use of L'Hopital's rule as excessive and recommends dividing both the numerator and denominator by the highest power of n to simplify the limit evaluation.

Areas of Agreement / Disagreement

Participants express differing views on the methods to determine convergence, with some favoring L'Hopital's rule while others advocate for simpler algebraic manipulation. There is no consensus on the preferred method or the overall approach to the problem.

Contextual Notes

Participants reference various mathematical techniques without resolving the underlying assumptions or limitations of each method. The discussion reflects differing interpretations of convergence tests and their applicability to sequences versus series.

Physicsisfun2005
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This seems like a simple problem...almost embarassed about it, but I can't figure it out:

Determine if the following sequence converges or diverges (2n-1)/(3n^2 +1), n= 1,2,3,... If the sequence converges, find its limit.


I think the series converges to 0...am i right? :confused:
 
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You mean the sequence converges to 0. Well, for one, why don't you just take the limit as n approaches infinity? Or, if you're not allowed to do that, show that the sequence only has positive numbers, and that it is decreasing (decreasing for all n > 2, in this case). Therefore it is bounded and monotone, hence convergent. Then assume that it converges to some small positive number e. Choose a natural number n > 2/(3e) and show that the sequence for this n is less than e. This gives a contradiction, showing that the sequence does not converge to any positive number, hence it must converge to 0.
 
sequence...opps yea......i did find the limit to get my answer using L' Hopital's rule...i was thinking this was the "nth term test" which cannot show convergence...only divergence...but then i realized this is a sequence and that test was for a series.
 
Last edited:
L'Hopital's rule seems like over-kill here. Standard method for dealing with fractions where n goes to infinity: Divide both numerator and denominator by the highest power of n (here n2) so that every term involves (1/n) to a power. As n goes to infinity, that goes to 0.
 

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