I find it hard to prove the convergence of $\sin(n^2)$ or $\frac{\tan(n)}{n^3}$ as $n$ goes to infinity. Is there any nice way to tackle this kind of limits?
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See e.g. here for discussion on $\sin\left(n^2\right)$: $\sin(n^2)$ diverges. – Minus One-Twelfth Apr 08 '19 at 11:48
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Questions such as this are related to measures of irrationality of $\pi$.
If $k \pi/2$ is the closest odd multiple of $\pi/2$ to $n$, we have $|\tan(n)| \sim 1/(|n - k \pi/2|)$, so in order to have $|\tan(n)|/n^3 > \epsilon > 0$ you'll want, for some $\delta > 0$, $|\pi - 2 n/k| < \delta/n^4$. It's almost certainly true that there are only finitely many such pairs $(n,k)$ (and therefore that $\tan(n)/n^3 \to 0$ as $n \to \infty$), but this has not been proven. As of 2012, the best known bound, due to Salikhov, is $7.606308$. So we can say that $\tan(n)/n^{6.606308} \to 0$, but $\tan(n)/n^3$ is not yet within reach.
Robert Israel
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