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Let’s say I have a sequence of real numbers, and I prove that the subsequence of even terms and the subsequence of odd terms both converge but not necessary to the same limit. Does that imply that the sequence converges? I am able to prove that if the above subsequences both converge to the same limit then the sequence also converges to that limit , but not quite sure about the above question. Thanks :)

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  • $\begingroup$ No in general. Consider the sequence $a_n=(-1)^n$. $\endgroup$ – Yadati Kiran Apr 19 at 18:16
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To conclude that the original sequence converges, we need that the even and odd subsequences converge to the same limit. Indeed, consider the sequence $a_n:=(-1)^n$. Then the even subsequence $a_{2n}=(-1)^{2n}=1$ converges to $1$ and the odd subsequence $a_{2n+1}=(-1)^{2n+1}=-1$ converges to $-1$. However, the sequence $a_n$ does not converge to a limit.

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  • $\begingroup$ So let’s say the sequence is increasing and I show that both even and odd subsequences are bounded ,that would mean they both have the same limit if and only if the original sequence also is bounded right? $\endgroup$ – Eden Hazard Apr 19 at 18:21
  • $\begingroup$ Thanks for your answer. :) $\endgroup$ – Eden Hazard Apr 19 at 18:21
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    $\begingroup$ If the original sequence is increasing and bounded above, then it has a limit, so the even and odd subsequences both converge to that same limit. Conversely, if both even and odd subsequences are bounded above and the original sequence is increasing, then the original sequence is bounded above and increasing so it converges and thus both even and odd subsequences converge to the same limit. $\endgroup$ – Dave Apr 19 at 18:28

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