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Let $$ a_{n} := \begin{cases} \sin(n) & \text{if n is odd}, \\ n & \text{if n is even}. \end{cases} $$

Does this sequence have a convergent subsequence? Justify your answer.

I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. I also know that $-1 < \sin(n) < 1$, but what about $n$ when $n$ is even?

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  • $\begingroup$ Is $a_n$ bounded at all? Where do you want to use Bolzano-Weierstrass Theorem? $\endgroup$ – SchrodingersCat Jan 9 '16 at 12:47
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    $\begingroup$ Hint: A subsequence of a subsequence of $(a_n)$ is also a subsequence of $(a_n)$. $\endgroup$ – Martin R Jan 9 '16 at 12:49
  • $\begingroup$ The B-W theorem tells you that the sequence $\left(\sin(2n+1)\right)_{n\in \mathbb N}$ has a convergent subsequence, i.e., there exists a strictly increasing map $\alpha\colon \mathbb N\to \mathbb R$ such that $\left(2\alpha(n)+1\right))_{n\in \mathbb N}$ is convergent. Is $\left(2\alpha(n)+1\right))_{n\in \mathbb N}$ a subsequence of $\left(a_n\right)_{n\in \mathbb N}$? Why or why not? $\endgroup$ – Git Gud Jan 9 '16 at 12:52
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    $\begingroup$ @Sophie Filer I noticed you have yet to accept any answers for any of the questions you've asked. Note that you can boost your reputation on this website by accepting an answer to your question, which you can do by clicking on the check mark underneath the downvote arrow on the answer you would like to accept. $\endgroup$ – Brandon Thomas Van Over Jan 9 '16 at 13:26
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As mentioned in the comments, a subsequence of a subsequence is a subsequence of the sequence. Since $\sin (2n+1)$ is bounded (as you mentioned) and it is a sequence in $\mathbb{R}^m$, Bolzano-Weierstrass tells us that it has a convergent subsequence. Hence this sequence has a convergent subsequence.

In the case where $n$ is even, we have a strictly increasing sequence which diverges. But we are only interested in finding a convergent subsequence for one of these (according to the question) so the answer above will suffice.

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