# 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?

• "so I can use Bolzano Weierstrass theorem to state that the sequence has a convergent subsequence." Yes. "Not sure where to go from here". You're not sure where to go from having reached an irrefutable conclusion? You don't have to go anywhere. You were ask whether it has a convergent subsequence and you answered that it did. So ... where else is there to go? – fleablood May 30 '18 at 18:39
• @fleablood I think the question asker was missing a subtlety for this question: $\sin(n)$ has a convergent subsequence, but it is not a subsequence of $(a_n)$ - the answer is, of course, to show that $\sin(n)$ for odd $n$ has a convergent subsequence, but the work presented in the question is not complete as it stands. – Milo Brandt May 30 '18 at 18:57
• It's not that subtle is it? $a_n$ where $n$ is odd s $a_n= \sin n$ is a subsequence. That subsequence is bounded so there is a subsequence of the subsequence that is bounded. A subsequence of a subsequence is a subsequence. You have to avoid being lazy and stupid ($a_n$ itself is not bounded and $a_n; odd$ is not $\{\sin n\}$ so you have to claim that $a_n; odd$ has the subsequence; not $\{\sin n\}$) but that's not hard to do. – fleablood May 31 '18 at 0:34
• "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." Okay, there are two observations you must make. $\{a_n\}$ is not bounded (but the subsequence $\{a_{2k+1}\} is$. And stating $\{\sin n\}$ has a convergent subsequence isn't enough as $\{\sin n\} \not \subset \{a_n\}$. But as $\{a_{2k+1} \}\subset \{\sin n\}$ is bounded there is a convergent $\{b_j\}\subset \{a_{2k+1}\} \subset \{\sin n\}$. That's fairly obvious and easy, but enough to make a question worth 5 marks. – fleablood May 31 '18 at 0:41

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 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.

You wrote (twice) “function” where you should have written “sequence”. Other than that, you are doing fine.

• Edited thanks. The question is worth 5 marks, I don't think this answer justifies this. Where do I go from here? Thanks. – Ben Jones May 30 '18 at 18:36
• You know that the sequence $(a_{2n-1})_{n\in\mathbb N}$ is bounded. Therefore, it has a convergent subsequence, which will be a subsequence of the original sequence, of course. – José Carlos Santos May 30 '18 at 18:37
• "Where do I go from here?" Well, you could go to breakfast. Or Tahiti, I hear Tahiti is nice this time of year. ... you've answered the question. You don't need to go anywhere. – fleablood May 30 '18 at 18:42
• I'm off to Tahiti then. – Ben Jones May 31 '18 at 18:36