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If $\{x_n\}$ is convergent and $\{x_ny_n\}$ is divergent, then $\{y_n\}$ is divergent.

Intuitively I believe this is true, since a divergent sequence must have a divergent subsequence. However, I have been reading about cauchy sequences which I dont claim to understand it fully seems to make me doubt my intuitive answer further.

I would appreciate it if someone can shed some light, thank you.

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  • $\begingroup$ By divergent do you mean 'non-convergent' or do you mean 'goes to infinity or to minus infinity'? These are not the same. $\endgroup$ – Fimpellizieri Apr 17 '18 at 0:55
  • $\begingroup$ @Fimpellizieri By convergent manual states that there there exist a limit l which the sequence converges to, for divergent the when this is not the case. So I guess, non-convergent. Hows that relate? I am not following. $\endgroup$ – C. Ekinci Apr 17 '18 at 1:39
  • $\begingroup$ If divergent is taken to mean 'not convergent', then the various proofs by contradiction below make sense. If divergent is taken to mean (as sometimes it is) 'divergent to infinity', then more care is needed. For instance, $(-1)^n$ is not convergent, but does not diverge to infinity. $\endgroup$ – Fimpellizieri Apr 17 '18 at 2:11
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Assume the contrary that $\{y_{n}\}$ is not divergent, then $\{y_{n}\}$ is convergent, then the product $\{x_{n}y_{n}\}$ is convergent since $\{x_{n}\}$ is convergent:

$|x_{n}y_{n}-xy|\leq|x_{n}-x||y_{n}|+|x||y_{n}-y|\leq(\sup_{n}|y_{n}|)|x_{n}-x|+|x||y_{n}-y|$.

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  • $\begingroup$ Ah, this is the product property of limits I believe? $\endgroup$ – C. Ekinci Apr 17 '18 at 0:50
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    $\begingroup$ Yes, you are right. $\endgroup$ – user284331 Apr 17 '18 at 0:51
  • $\begingroup$ Thank you very much! I considered contradiction but not that it would open up to this. $\endgroup$ – C. Ekinci Apr 17 '18 at 0:52
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Your intuition is correct! You can prove this by contradiction: Suppose that $y_n$ is convergent. Since $x_n$ is also convergent, the sequence $x_n y_n$ converges, a contradiction.

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  • $\begingroup$ Thank you, I as I said above I did consider contradiction but I didnt put the two together. $\endgroup$ – C. Ekinci Apr 17 '18 at 0:53
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If not then you have two convergent sequences and the product would be convergent. That contradicts your assumption.

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