# Does $a_{n}$ have a convergent subsequence

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

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?
• Is $a_n$ bounded at all? Where do you want to use Bolzano-Weierstrass Theorem? Jan 9 '16 at 12:47
• Hint: A subsequence of a subsequence of $(a_n)$ is also a subsequence of $(a_n)$. Jan 9 '16 at 12:49
• 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? Jan 9 '16 at 12:52
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.