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Sequence limits?

  • Thread starter dannysaf
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  • #1
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Let lim n →∞ XnYn = 0. Is it true that Limn →∞ Xn= 0
or Limn →∞Yn = 0 (or both)?
 

Answers and Replies

  • #2
216
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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:
 
  • #3
VietDao29
Homework Helper
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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?
 
  • #4
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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.
 
  • #5
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Yup, sorry I forgot to mention that the series must converge, in other way they would be divergent and we could not separate them.
 
  • #6
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Simple counterexample. Clearly, [tex]\lim 1/n = \lim \(n \cdot 1/n^2 \) = 0[/tex] but we have that [tex]\lim n = +\infty[/tex] and [tex]\lim 1/n^2 = 0[/tex].
 
  • #7
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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
 
  • #8
VietDao29
Homework Helper
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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:
 
  • #9
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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:

Simple counterexample. Clearly, [tex]\lim 1/n = \lim \(n \cdot 1/n^2 \) = 0[/tex] but we have that [tex]\lim n = +\infty[/tex] and [tex]\lim 1/n^2 = 0[/tex].
To which my original comment makes more sense. Sorry for any confusion.

--Elucidus
 
  • #10
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Let [itex]x_n [/itex] be alternately 1 and 0, let [itex]y_n [/itex] be alternately 0 and 1.
Then [itex]x_n y_n = 0 [/itex] for all n, but neither individual limit exists. In particular, neither individual limit is zero.
 
  • #11
HallsofIvy
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Let [itex]x_n [/itex] be alternately 1 and 0, let [itex]y_n [/itex] be alternately 0 and 1.
Then [itex]x_n y_n = 0 [/itex] for all n, but neither individual limit exists. In particular, neither individual limit is zero.
Excellent example. However, neither [itex]x_n[/itex] nor [itex]y_n[/itex] 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
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Excellent example. However, neither [itex]x_n[/itex] nor [itex]y_n[/itex] 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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