Seq Limits: Xn, Yn & Lim n→∞ = 0?

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Homework Help Overview

The discussion revolves around the limits of sequences Xn and Yn, particularly in the context of their product approaching zero as n approaches infinity. Participants explore whether this implies that at least one of the sequences must also converge to zero.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants examine the conditions under which the limit of the product of two sequences equals zero, questioning the necessity of the individual limits existing. Some raise concerns about cases where sequences oscillate or do not converge.

Discussion Status

The discussion is active, with various counterexamples being presented to challenge the original proposition. Some participants suggest that the original statement may not hold true under certain conditions, while others clarify the need for convergence in the sequences for the limit product to be valid.

Contextual Notes

There is an emphasis on the convergence of the sequences, with some participants noting that without convergence, the limits cannot be separated. Additionally, the discussion includes examples where the product of sequences approaches zero while the individual sequences do not converge to zero.

dannysaf
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Let lim n →∞ XnYn = 0. Is it true that Limn →∞ Xn= 0
or Limn →∞Yn = 0 (or both)?
 
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lim n →∞ XnYn = lim n →∞ Xn * lim n →∞ Yn

One or even both limits need to be 0 so that lim n →∞ Xn * Yn =0

I guess you are right. :smile:
 
njama said:
lim n →∞ XnYn = lim n →∞ Xn * lim n →∞ Yn

Well, this is only true when both xn, and yn have limits. So, what if they don't? :wink: Say, what if they're oscillating?
 
njama said:
lim n →∞ XnYn = lim n →∞ Xn * lim n →∞ Yn


This equation is only correct if the individual limits on the right-hand side exist. The proposition in the original post is false and can be disproven with a counterexample.
 
Yup, sorry I forgot to mention that the series must converge, in other way they would be divergent and we could not separate them.
 
Simple counterexample. Clearly, \lim 1/n = \lim \(n \cdot 1/n^2 \) = 0 but we have that \lim n = +\infty and \lim 1/n^2 = 0.
 
VietDao29 said:
Well, this is only true when both xn, and yn have limits. So, what if they don't? :wink: Say, what if they're oscillating?

This example satisfies the original claim that at least one of the sequences converges to 0. A counter example for this must consist of two sequences whose product vanishes but both sequence do not converge to 0.

--Elucidus
 
Elucidus said:
This example satisfies the original claim that at least one of the sequences converges to 0. A counter example for this must consist of two sequences whose product vanishes but both sequence do not converge to 0.

--Elucidus

You've misquoted yet again, Elucidus.. :cry: :cry:

You shouldn't forget to wear your glasses, then.. :wink:
 
VietDao29 said:
You've misquoted yet again, Elucidus.. :cry: :cry:

You shouldn't forget to wear your glasses, then.. :wink:

Indeed. I'm surprised. I was trying to quote this post:

fmam3 said:
Simple counterexample. Clearly, \lim 1/n = \lim \(n \cdot 1/n^2 \) = 0 but we have that \lim n = +\infty and \lim 1/n^2 = 0.

To which my original comment makes more sense. Sorry for any confusion.

--Elucidus
 
  • #10
Let x_n be alternately 1 and 0, let y_n be alternately 0 and 1.
Then x_n y_n = 0 for all n, but neither individual limit exists. In particular, neither individual limit is zero.
 
  • #11
g_edgar said:
Let x_n be alternately 1 and 0, let y_n be alternately 0 and 1.
Then x_n y_n = 0 for all n, but neither individual limit exists. In particular, neither individual limit is zero.
Excellent example. However, neither x_n nor y_n converges and the O.P. finally told us.
Yup, sorry I forgot to mention that the series must converge, in other way they would be divergent and we could not separate them.
 
  • #12
HallsofIvy said:
Excellent example. However, neither x_n nor y_n converges and the O.P. finally told us.

HallsofIvy I am not the O.P, mate. :biggrin: I was just adding, additional explanation of my first post. :cool:
 

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