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I want to prove that the $\limsup\limits_{n\rightarrow \infty} \sin n=1$. I know that $1$ is an upper bound for $\sin n$ but I cannot find a subsequence of $\sin n$ that converges to $1$. Can somebody help me construct such a subsequence?

Kola B.
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OKPALA MMADUABUCHI
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    $\sin\left(n\pi+\frac{\pi}{2}\right)$? – cand May 11 '15 at 00:28
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    http://math.stackexchange.com/questions/63526/showing-sup-sin-n-mid-n-in-mathbb-n-1 – ThePortakal May 11 '15 at 00:29
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    @cand sorry but that is not a subsequence of ${\sin n}$. – Tim Raczkowski May 11 '15 at 00:30
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    @cand I think he meant $n \in \mathbb N$ – ThePortakal May 11 '15 at 00:30
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    Can't you just use the ergodic theorem on the circle (which is easy to prove directly) to say that there are integers that get arbitrarily close to $\frac{\pi}{2}$ (working modulo $2\pi$, of course)? – shalop May 11 '15 at 00:34
  • A sketch of a direct method: find a sequence of positive integers $x_n$ and a sequence of positive integers $y_n$ which are equivalent to $1$ mod $4$ such that $q_n = x_n/y_n \to \pi/2$. Then certainly $\sin(q_n) \to 1$. Try to make the convergence sufficiently rapid as $y_n$ grows that $\sin(x_n) = \sin(y_n q_n) \to 1$. This "make the convergence rapid as $y_n$ grows" idea leads naturally into continued fraction expansion. – Ian May 11 '15 at 15:02
  • https://math.stackexchange.com/questions/4764/sine-function-dense-in-1-1 – Guy Fsone Nov 28 '17 at 18:32

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There is a theorem by Kronecker (?) that says that the grid in $\Bbb R$ generated by $1$ and some number $r$ is either dense or an arithmetic progression, obviously corresponding to the cases $r$ irrational or rational.

In this case it means that the sequence of integers modulo $2\pi$ is dense in the interval $[0,2\pi)$, and thus the set of values $\sin(n)$ dense in the interval $[-1,1]$.

Lutz Lehmann
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