Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence? Does the sequence $(a_n)_{n\in\mathbb{N}}$ have a convergent subsequence?
$$a_n= \begin{cases} \sin(n), & \text{if $n$ is odd} \\ n, & \text{if $n$ is even} \end{cases}$$
I understand that $\sin(n)$ is a bounded sequence so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence. Not sure where to go from here assuming I am along the right lines?
 A: Let's construct the subsequence which converges to $0$.
Consider convergents of $\pi$ continued fraction: $\left\{ \dfrac{n_j}{d_j},\; j\in\mathbb{N} \right\}$. And focus on its numerators:
$$n_1 = 3,\\ n_2 = 22, \\ n_3 = 333, \\ n_4 = 355, \\ n_5 = 103993, \\ n_6 = 104348, \\ \vdots $$
Then $$\lim_{j\to \infty} \sin n_j =0.$$
Indeed, easy to estimate: $$|\sin n_j| = \left| \sin\left( n_j - \pi d_j  \right) \right| \approx \left| n_j - \pi d_j \right|=d_j\left| \dfrac{n_j}{d_j} - \pi \right| < \dfrac{1}{d_{j+1}}. $$
And it remains to show that the sequence $\{n_j, \; j\in\mathbb{N}\}$ contains infinite number of odd $n_j$.
If continued fraction (of $\pi$) denote as  $[a_0;a_1, a_2, \ldots, a_j, \ldots]$, then
$$
n_{j+1} = a_{j+1}n_j + n_{j-1}.
$$
There are $8$ possibilities:
\begin{array}{|l|l|l|l|}
\hline 
(n_{j-1}, n_j) & a_{j+1} & n_{j+1} & \rightarrow (n_j, n_{j+1})\\
\hline 
(odd, odd) & odd & even & \rightarrow (odd, even) \\
(odd, odd) & even & odd & \rightarrow (odd, odd) \\
\hline
(odd, even) & odd & odd & \rightarrow (even, odd) \\
(odd, even) & even & odd & \rightarrow (even, odd) \\
\hline
(even, odd) & odd & odd & \rightarrow (odd, odd) \\
(even, odd) & even & even & \rightarrow (odd, even) \\
\hline
(even, even) & odd & even & \rightarrow (even, even) \\
(even, even) & even & even & \rightarrow (even, even) \\
\hline
\end{array}
which show that if at least one of $n_j$ is odd, then there is infinite number of odd $n_j$.
A: You wrote (twice) “function” where you should have written “sequence”. Other than that, you are doing fine.
A: A subsequence of a subsequence is a subsequence. The sequence of values of $a_n$ for which $n$ is odd is a subsequence, and you've already said that that has a convergent subsequence. That convergent subsequence is a subsequence of a subsequence of the original sequence. Therefore it is a subsequence of the original sequence.
