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Can we prove that $\displaystyle \limsup_{n\to \infty} \sin(n) = 1$?

I can prove that the above statement holds assuming that $\displaystyle \frac{\pi}{2}$ is normal (this fact is used somewhat tangentially in the proof, making me wonder whether it can be proved without it): Fix $\epsilon > 0$ and find $5^n > \frac{1}{\epsilon}$. We may find a string $00\dots 01$ (in base $5$) with $n$ zeroes occuring at the $m^{th}$ digit and it follows that:

$$\frac{\lfloor \frac{\pi}{2} \cdot 5^m \rfloor}{5^m} < \frac{\pi}{2} < \frac{\lfloor \frac{\pi}{2} \cdot 5^m \rfloor + \epsilon}{5^m}$$

So $d(\frac{\pi}{2} \cdot 5^m , \lfloor \frac{\pi}{2} \cdot 5^m \rfloor ) < \epsilon$. Since $5^m \equiv 1 \mod 4$, we have $\sin(\frac{\pi}{2} \cdot 5^m) = 1$, and since $|\sin|$ dominates the triangle wave, $d(1,\sin(\lfloor \frac{\pi}{2} \cdot 5^m \rfloor ) ) < \displaystyle \frac{\epsilon}{\pi}$, from which the result follows.

But $\displaystyle \frac{\pi}{2}$ is not proven to be normal. Can we prove this fact without using normality?

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The more general fact that $\{\sin(n)\mid n\in\mathbb N\}$ is dense in $[-1,1]$ is proved here: math.stackexchange.com/questions/4764/… –  Andres Caicedo Apr 12 '13 at 5:30

1 Answer 1

up vote 4 down vote accepted

Normality is an overkill indeed ; irrationality suffices.

Since $2\pi$ is irrational, the additive subgroup ${\mathbb Z}-{2\pi}{\mathbb Z}$ is dense in $\mathbb R$. In fact, ${\mathbb N}-{2\pi}{\mathbb N}$ is also dense in $\mathbb R$ (this is well-known and not difficult to show ; use the pigeon-hole principle).

So for any $\varepsilon \gt 0$, there are positive integers $n,k$ with $\big|n-(2\pi)k-\frac{\pi}{2}\big| \lt \varepsilon$, and hence $|\sin(n)-1| \lt \varepsilon$.

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