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I'm trying to prove that $\limsup \sin(n) = 1$. This fact sounds obvious to me, but I don't know how to use the given hint that any numbers of the form $n + 2m\pi$, $m,n \in \mathbb{Z}$, are dense in the real numbers. (I also can't figure out how to prove this, aside from taking $a < b$, defining the positive difference $b - a$, and using density of the irrationals.)

Any hints or help would be appreciated.

amWhy
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  • That helps for proving denseness. I will try to add an updated attempt for that. But it doesn't fully answer my question, because I still want to prove $\lim sup \sin n = 1$. –  Feb 16 '21 at 19:35
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    $[-1,1]$ is the set of limit points. $\limsup$ is the biggest limit point. – Tito Eliatron Feb 16 '21 at 19:42
  • @StanleySmith The number of the form $n+2m\pi$ are NOT dense in $\mathbb{R}$. And the result is not obvious at all... what is obvious is that $\limsup (\sin n ) \leq 1$. – PierreCarre Feb 17 '21 at 16:58

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The superior limit of a sequence determines the limit of the largest convergent subsequence. Let $$S=\{\sin(n): n\in \mathbb{Z}\}.$$ We already know that that $\sin(x)$ is upper-bounded by $1$, and if we consider the sequence $$\tilde{S}=\{\sin(\pi n):n\in \mathbb{Z}\},$$ we can clearly see that $$\lim\sup \tilde{S}=1.$$ The main issue is that $$\{\pi n:n\in\mathbb{Z}\}\not\subset\mathbb{Z}\quad\forall n.$$ This is where the hint is needed. Although the points in $\{\pi n:n\in \mathbb{Z}\}$ do not belong in $\mathbb{Z}$ (except for $n=0$), thanks to $\{n+m\pi:n,m\in\mathbb{Z}\}$ being dense in $\mathbb{Z}$, given $n$, we can find $\tilde{n}, m\in \mathbb{Z}$ so that $\tilde{n}$ is as close as we want to $m+\pi n$ $\forall n$.

  • Is $sin(n\pi) = 0$ for any n? If so, $\limsup{\overline S}=0$ as well. – Salcio Feb 17 '21 at 10:58
  • $\sin(n\pi)=0$ for $n=0$, but that does not imply that $\lim\sup \tilde{S}=0$. We need 'infinitely' many $n$'s to satisfy either $\sin(n\pi)=0$ or $|\sin(n\pi)|<\epsilon$ for arbitrarily small $\epsilon$. – Arroz con Tomate Feb 17 '21 at 12:06