Intuition behind a convergent subsequence of $\sin(n)$

Question

$\let\eps\varepsilon$

I was looking through a Real Analysis exam paper one day and was stuck on a question; fortunately there is a solution provided which I will sketch below, but I have no intuition on how I would have come up with this proof.

Question: Find an explicit convergent subsequence of $\sin(n)$ and prove it converges.

My first guess was to try $ \sin\lfloor 2\pi n\rfloor $ (or ceiling instead of floor) and hope $\lfloor 2\pi n \rfloor$ will approximate some multiple of $2\pi$ reasonably well for large $n$. However, on reflection this probably won't work because a subsequence of this is $\sin\lfloor 2\pi \cdot 10^n \rfloor $, and (ignoring sign) $2\pi \cdot 10^n -\lfloor 2\pi \cdot 10^n \rfloor$ is the fractional part of $2\pi\cdot 10^n$ which I have no idea how to control, much less inside of a $\sin$.

After this I was stuck for a good number of hours and eventually gave up. The solution given, is amazingly similar to my attempt: it uses the subsequence $$a_n := \sin \sqrt{\left\lceil (2\pi n)^2 \right\rceil} \equiv \sin b_n$$ It is perhaps at this point I should note that the question did say to instead find a subsequence of $\sin\sqrt{n}$ instead; but I thought I was simplifying the question as $\sin(n) = \sin\sqrt{n^2}$ is a subsequence of $\sin \sqrt{n}$.

Anyway, the answer drops out after a little work. Noting that we can write $$b_n = 2\pi n + \eps_n, \quad \eps_n\in [0,1) $$ and making the easy approximation (which was given as a hint) $|\sin x | \leq |x|$, one obtains $$|\sin b_n| = \left|\sin \left[\sqrt{\left\lceil (2\pi n)^2 \right\rceil} - 2\pi n\right]\right| \leq 2\pi \left|n\sqrt{1+\frac{\eps_n}{4\pi^2n^2}} - n\right| $$

And by Bernoulli's inequality, $(1+\eps)^n \geq 1+n\eps$, applied with $n=2$, $$ \sqrt{1+\frac{\eps_n}{4\pi^2}} \leq 1 + \frac{\eps_n}{8\pi^2n^2} \xrightarrow[n\to\infty]{} 1$$

Thus $|\sin b_n|$ is bounded by a null sequence; by the Sandwich Rule it is itself null, which concludes the proof.

Notably, trying the above argument for $\sin\lceil 2\pi n \rceil$ gives you the very useless $$|\sin(2\pi n + \eps_n) | = | \sin(2\pi n + \eps_n - 2\pi n) | \leq \eps_n$$

How did the author of the exam come up with this solution? Is there anything intuitive in retrospect you can share? (I would appreciate more answers that don't say 'the exam question was phrased in a way to hint to the expected answer', because I find this a little unsatisfactory)

Thanks!

Answer 1

Here's an attempt to provide intuition as to why we can work with $\sin(\sqrt x)$ in a way that we can't with $\sin(x)$:

• The equidistribution theorem tells us that $2 \pi n - \lfloor 2 \pi n \rfloor$ will stay "evenly" spread out between $0$ and $1$. On average, it will "miss the mark" of $0$ by $1/2$. Not particularly useful in producing an upper bound for this difference, but a good explanation of how "uncontrollable" this sequence is
• The function $f(x) = \sin(\sqrt{x})$ becomes "more spread-out" as $x \to \infty$ (that is, $f'(x) \to 0$). As a result, if we can get a sequence $x_n$ such that $|x_n - (2 \pi n)^2| \leq 1$, we will find that $$ \lim_{n \to \infty} \left|f([2 \pi n]^2) - f(x_n)\right| = \lim_{n \to \infty} f'(c_n)\left|(2 \pi n)^2 - x_n\right| \to 0 $$ Where the appropriate $c_n$ is chosen between $2 \pi n$ and $x_n$. Such a $c_n$ is guaranteed to exist by the mean value theorem.
• The function $\sin(x)$ does not exhibit this behavior