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I'm solving an exercise from The elements of Real Analysis by Robert G. Bartle, which asks to find the $\limsup$ and $\liminf$ of the sequence given by $a_n = n (\sin n)^2$.

I have already calculated $\limsup a_n$ as follows: Since $\left\lbrace\sin n : n\in\mathbb{N}\right\rbrace$ is dense in $[-1, 1]$, and $\sin n \neq 1$ for every $n\in\mathbb{N}$, there are infinite many numbers $m\in\mathbb{N}$ such that $\sin m \geq \frac{9}{10}$, and therefore, $m(\sin m)^2\geq \frac{81}{100}m$ for infinite many natural numbers. This implies $a_n$ is not bounded above, so $\limsup a_n = +\infty$.

I was trying to apply something similar to find $\liminf a_n$, but haven't been able to get any conclusion. Can you help me with that, please?

Iván G M
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  • Are you familiar with results on rational approximations like Hurwitz's theorem, Roth's theorem, Liouville's theorem, etc.? – anomaly Apr 09 '24 at 16:17
  • Not really, at this point the book hasn't covered so many advanced concepts like those. – Iván G M Apr 09 '24 at 16:20
  • It is $0$ because $\pi$ is irrational and it means that there are integers $p,q$ such that $|\pi - p/q| < 1/q^2$. The last inequality can be proved using "pigeon hole principle". – Salcio Apr 09 '24 at 17:59
  • Do you mind explaining how to proceed, please? I think we can get $\left\lvert \sin(q\pi)-\sin(p)\right\rvert<\left\lvert q\pi-p\right\rvert < 1/q$, so $p(\sin p)^2<p/{q^2}$. Am I on the right track? – Iván G M Apr 09 '24 at 18:21
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    For small $x$ sine is of the order $x$. Also you messed up $p$ and $q$ I think. We know that $|q\pi -p|$ is small and that $sin {q\pi} = 0$. This should do it. – Salcio Apr 09 '24 at 18:38
  • Okay, got it, thank you! – Iván G M Apr 09 '24 at 18:40
  • It may not be quite as simple as that: $n(\sin n)^2 \approx 3.226 \times 10^{-7}$ when $n=355$ but I think the next $n$ which gives a lower value may be $n=5419351$. – Henry Apr 10 '24 at 00:10

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