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I have looked through previous posts but have been struggling with this problem.

The sequence is {$a_n$} and its subsequences {$a_{2k}$}, {$a_{2k+1}$}, {$a_{3k}$} converge. I have to prove that {$a_n$} converges.

I know that a sequence converges if all of its subsequences converge. I'm suspecting that I have to prove that every subsequence belongs into these 3 subsequences. Thank you for your time and help.

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If all of the even subsequences converge and all of the odd subsequences converge, then the original sequence will converge iff both of the above subsequences converge to the same value.

To show both the even and odd subsequences converge to the same value, you need a third subsequence that converges and covers more than a finite amount of even and odd values. That third subsequence is given here as $\{a_{3k}\}$.

More generally, if two subsequences converge and every term in the original sequence belongs to one of the two subsequences, then convergence of the original sequence requires a third sequence that takes terms from both sequences infinitely many times.

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  • $\begingroup$ Quick question. Can I just assume that every even and odd subsequences are converging to $a$? I am not understanding how {$a_{3k}$} contains subsequences from both even and odd subsequences - let alone, how it plays an integral part to this. Resources I'm using (pirate.shu.edu/~wachsmut/ira/numseq/proofs/subconv.html) and (math.stackexchange.com/questions/539964/…). I just said that for {$a_j$} approaches $a$ when $n$>$N_1$ and $j$ is odd. Likewise {$_j$} approaches $a$ when $n$>$N_2$ and $j$ is even. $\endgroup$ – ozarka Sep 28 '16 at 23:41
  • $\begingroup$ And I showed that {$a_n$} approaches $a$ when $n$>$M$ where $M$=max{$N_1$,$N_2$}. $\endgroup$ – ozarka Sep 28 '16 at 23:46
  • $\begingroup$ @ozarka Can $3k$ result in an odd number as $k\to\infty$? Can it result in an even number? So if $a_{3k}$ converges to something, the even parts of $a_{3k}$ and the odd parts of $a_{3k}$ also converge to something. Clearly, they converge to whatever you say $a_{3k}$ converges to, but they also converge to what the even/odd subsequence converges $(a_{2k},a_{2k+1})$ to. Does that make more sense? $\endgroup$ – Simply Beautiful Art Sep 28 '16 at 23:47
  • $\begingroup$ Yes! So I'll assumed {$a_3k$} converges to $a$. I see that $a_6$ is a subsequence of both $a_{3k}$ and $a_{2k}$. I also see that $a_9$ is a subsequence for both $a_{3k}$ and $a_{2k+1}$. Because $a_3k$ is converging to $a$, surely it's subsequences are also (?). This means that $a_{2k}$ and $a_{2k+1}$ is also approaching $a$. Is this the correct thought process? $\endgroup$ – ozarka Sep 28 '16 at 23:51
  • $\begingroup$ @ozarka Yup! You got it ;) (btw, use \{ and \} to make nice braces $\endgroup$ – Simply Beautiful Art Sep 28 '16 at 23:52
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Hint: it is sufficient to show that the convergent sequences $\{a_{2k}\}$ and $\{a_{2k+1}\}$ have the same limit. To see that they do have the same limit, note that $\{a_{2k}\}$ and $\{a_{2k+1}\}$ each have a subsequence that is a subsequence of the convergent sequence $\{a_{3k}\}$.

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  • $\begingroup$ This was super helpful. Thank you. $\endgroup$ – ozarka Sep 28 '16 at 22:08
  • $\begingroup$ Quick question. Can I just assume that every even and odd subsequences are converging to $a$? I am not understanding how {$a_{3k}$} contains subsequences from both even and odd subsequences - let alone, how it plays an integral part to this. Resources I'm using (pirate.shu.edu/~wachsmut/ira/numseq/proofs/subconv.html) and (math.stackexchange.com/questions/539964/…). I just said that for {$a_j$} approaches $a$ when $n$>$N_1$ and $j$ is odd. Likewise {$_j$} approaches $a$ when $n$>$N_2$ and $j$ is even. $\endgroup$ – ozarka Sep 28 '16 at 23:45
  • $\begingroup$ And I showed that {$a_n$} approaches $a$ when $n$>$M$ where $M$=max{$N_1$,$N_2$}. $\endgroup$ – ozarka Sep 28 '16 at 23:46

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