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

[Physicsisfun2005]

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?

 

[AKG]

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.

 

[Physicsisfun2005]

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.

 

[HallsofIvy]

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.