Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The more general fact that $\{\sin(n)\mid n\in\mathbb N\}$ is dense in $[-1,1]$ is proved here:… – Andrés E. Caicedo Apr 12 '13 at 5:30
up vote 5 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 $|\sin(n)-1| \lt \varepsilon$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.