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I am looking for a (simple) formula for an increasing integer sequence $a_n$ such that $$\sin(a_n)\to 0.$$

Of course such a sequence exists, but I am hoping for a simple formula for such a sequence that I could show my students.

The context is the following: I wrote $a_n=\sin(n)$ on the board and my students confidently declared that it wasn't convergent. Fair enough...so I wrote $b_n=\sin(n^2)$ on the board. Again, they were confident it wasn't convergent. I explained to them that they had the tools to show $a_n$ was not convergent, but showing $b_n$ was not convergent would take a lot more work and wasn't at all obvious. So, I'm hoping for a counterexample to show them that it really isn't obvious whether a sequence $\sin(q_n)$ converges as $n\to\infty$. (A simple sequence where the answer is unknown by researchers would also be interesting.)

Jason Siefken
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  • @Semiclassical Of course, $\pi n$ is not an integer. The other obvious one, $a_n = 0$ is not increasing. – GEdgar Mar 15 '23 at 16:46
  • It is not clear from your phrasing whether you think $\sin n^2$ is convergent or not. "but showing bn was convergent would take a lot more work and wasn't at all obvious." – coffeemath Mar 15 '23 at 16:46
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    It is "not at all obvious" indeed! it is false. – Anne Bauval Mar 15 '23 at 16:49
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    Does this answer your question? Subsequence of $\sin n$ – Anne Bauval Mar 15 '23 at 16:53
  • @coffeemath, Anne Bauval, thank you for pointing out the error. I meant to say that showing b_n was not convergent took more work. – Jason Siefken Mar 15 '23 at 21:43
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    Why do you think that there exist an simple formula for such sequence? – Diego Santos Mar 15 '23 at 22:52
  • What is needed can be shown also using the theory of Farey sequences and the fact that $\pi$ is irrational. – coffeemath Mar 15 '23 at 23:04
  • @DiegoSantos I don't know it for a fact, and "simple" is subjective, but formulas like https://en.wikipedia.org/wiki/William_Brouncker,_2nd_Viscount_Brouncker#Brouncker's_formula exist, which makes me guess there's a satisfying answer out there somewhere. – Jason Siefken Mar 16 '23 at 02:58
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    Not integers, but will rational multiples of $\pi$ work? If so, then students should readily see that $a_n = (\pi/2)(1-2^n)$ is increasing, but also should see that it $\sin(a_n)$ converges to $0$. OTOH, if you need $a_n$ to not converge... that's a bit harder. – Eric Snyder Mar 16 '23 at 09:02
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    @EricSnyder - not much harder. Just add $2\pi n$. @ Jason Siefken - You can indeed use continued fractions to find approximations to $\pi$ allowing you to use the denominators as your sequence. But your students are not going to find that sequence - no matter how much you try to pretty it up - to be "simple". – Paul Sinclair Mar 16 '23 at 18:49
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    @PaulSinclair Exactly! That's why I asked the question :-) I am not aware of a "nice" formula for those denominators. – Jason Siefken Mar 16 '23 at 22:32

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