Let lim n →∞ XnYn = 0. Is it true that Limn →∞ Xn= 0
or Limn →∞Yn = 0 (or both)?
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.
Well, this is only true when both xn, and yn have limits. So, what if they don't? Say, what if they're oscillating?
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, [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].
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.
You've misquoted yet again, Elucidus..
You shouldn't forget to wear your glasses, then..
Indeed. I'm surprised. I was trying to quote this post:
To which my original comment makes more sense. Sorry for any confusion.
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.
HallsofIvy I am not the O.P, mate. I was just adding, additional explanation of my first post.
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