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$|a_n|$ is diverging.
How do you prove $a_n$ (without absolute value) has a subsequence converging to a finite limit?

I know that if a sequence has two subsequences converging to different numbers, then the sequence is diverging. May I use the opposite lemma?

EDIT:
The sequence $|a_n|$ doesn't diverging to $\infty$. So you may say the sequence has no limit.

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    $\begingroup$ thats false, take $a_n$ positives to begin with. $\endgroup$ – azarel Nov 21 '13 at 21:21
  • $\begingroup$ With the edit it still is false: $\;a_n=(-1)^nn\;$ $\endgroup$ – DonAntonio Nov 21 '13 at 21:41
  • $\begingroup$ Your example is wrong because $|a_n|$ can't diverge to infinity $\endgroup$ – Daniel Gagnon Nov 21 '13 at 21:56
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That's not true: Take the sequence $$a_n = n$$ Then every subsequence diverges.

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  • $\begingroup$ I have edited my question. Please have a look $\endgroup$ – Daniel Gagnon Nov 21 '13 at 21:29
  • $\begingroup$ @DanielGagnon So is your sequence divergent, but bounded? Are you familiar with the Bolzano-Weierstrass Theorem? $\endgroup$ – user61527 Nov 21 '13 at 21:31
  • $\begingroup$ Let me interpret what you're saying: if the sequence has no limit and doesn't diverge to $\infty$ then it must be bounded, and therefore must have a subsequence with a finite limit according to BW Theorem. Right? $\endgroup$ – Daniel Gagnon Nov 21 '13 at 21:41
  • $\begingroup$ @DanielGagnon If by "doesn't diverge to $\infty$" you mean that it's bounded, then yes. $\endgroup$ – user61527 Nov 21 '13 at 21:42
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    $\begingroup$ @Daniel: No! It can be unbounded, but still not converge to infinity. For instance, if $a_{2n} = n$ but $a_{2n+1} = 1$. $\endgroup$ – TonyK Nov 29 '13 at 9:38

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