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It is known that $\{\sin n : n\in\mathbb{N}\}$ is dense in $[-1,1]$, hence $\lim_{n\to\infty}\sin n$ doesn't exist and also $\lim_{n\to\infty} n^t\sin n$ doesn't exist for all $t>0$ (the reason is that the density implies that inequalities $\sin n>\frac{1}{2}$ and $\sin n<-\frac{1}{2}$ are satisfied infinitely many times, so there are subsequences tending to $+\infty$ and $-\infty$).

What about $\lim_{n\to\infty} |n^t\sin n|$ ?

The above argument shows that the limit - if exists - is infinite

I dont't think it does converge, but I don't know how to prove it.

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  • $\begingroup$ A divergent sequence does not have a limit. But we can say (in this case) to which extreme it diverges too. This limit diverges to $+\infty$ (think about what it is you're really trying to show here). $\endgroup$ Mar 13 '16 at 23:53
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    $\begingroup$ If $t \leqslant 1$, it doesn't tend to $+\infty$: for the sequence $\{n_m\}_{m=1}^{\infty}$ of numerators of best rational approximations of $\pi$ (from Dirichlet's theorem) we have that $\lvert n_m \sin n_m \rvert$ is bounded. I have no idea, how to deal with $t > 1$, though. $\endgroup$ Mar 14 '16 at 0:29
  • $\begingroup$ You're right about $t\leq 1$, but when $t>1$ it's actually fairly easy to argue divergence, since the $n^t$ factor blows up (and the $\sin n$ factor does not tend to $0$). $\endgroup$ Mar 14 '16 at 1:22
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    $\begingroup$ @JustinBenfield A sequence $x_n:\Bbb N\to \Bbb R\cup \{\pm\infty\}, x_n\to +\infty$ does converge, though. $\endgroup$ Mar 14 '16 at 3:12
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The question is strongly connected with irrationality measure of $\pi$. That is such number $\mu$, that for every numbers $\lambda, \nu$ with $\lambda < \mu < \nu$:

  1. there exist infinitely many distinct rational numbers $p/q$ for which $$ \left| \frac{p}{q} - \pi \right| < \frac{1}{q^\lambda} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert < q^{1-\lambda} $$

  2. for each rational $p/q$ with sufficiently large denominator $$ \left| \frac{p}{q} - \pi \right| > \frac{1}{q^\nu} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert > q^{1-\nu} $$

Exact value of $\mu$ is not known, but it's true that $2\leqslant \mu \leqslant 7.6063$. Note, that by Dirichlet's theorem, the first item is always true for $\lambda = 2$, disregarding the irrationality measure.

Returning to question: there is watershed for the parameter. If $t > \mu-1$, then the limit is $+\infty$; if $t < \mu-1$, then it doesn't exist. The remaining case $t = \mu-1$ depends on the behavior of diophantine approximations of $\pi$.

First case: $t < \mu-1$.

If $t < \mu-1$ and $p/q$ satisfies the first inequality, then $$ \lvert \sin p \rvert = \lvert \sin (p-\pi q) \rvert \leqslant \lvert p - \pi q \rvert \leqslant q^{-t} \sim \pi^t p^{-t} = O(p^{-t}) \;\text{ as }\; p\rightarrow \infty $$ hence there is bounded subsequence of $\{ n^t \sin n \}_{n=1}^\infty$ and it cannot have infinite limit.

Second case: $t > \mu-1$.

Take $\varepsilon > 0$, such that $t-\varepsilon > \mu-1$. Given $n\in \mathbb N$, choose $m \in \mathbb N$, such that $\lvert n -\pi m \rvert \leqslant \frac{\pi}{2}$. When $n$ is sufficiently large, $$ \lvert \sin n\rvert = \lvert \sin(n-\pi m) \rvert \geqslant \tfrac{2}{\pi} \lvert n - \pi m \rvert = \tfrac{2}{\pi} \lvert n - \pi m \rvert \geqslant \tfrac{2}{\pi} m^{-t+\varepsilon} \sim 2\pi^{t-\varepsilon-1} n^{-t+\varepsilon}, $$ hence $\lvert n^t \sin n \rvert \geqslant C n^\varepsilon \rightarrow +\infty$.

The remaining case: $t = \mu-1$.

There are two alternatives: $(A)$ whether exist $C > 0$ and infinitely many rational solutions $p/q$ of $$ \left| \frac{p}{q} - \pi \right| \leqslant \frac{C}{q^\mu} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert \leqslant C q^{1-\mu} = C q^{-t}$$ or $(B)$ the converse. As I already mentioned, if $\mu = 2$, then $(A)$ holds.

Suppose $(A)$ is true. Then by the same argument, as in the first case, $\nexists\lim\limits_{n\rightarrow\infty} \lvert n^t \sin n \rvert$.

Oppositely, we have $(B)$, which, in fact, is equivalent that for every sequence of distinct rationals $\{p_n / q_n \}_{n=1}^\infty$ the sequence $\{ q_n^t \lvert p_n - \pi q_n \rvert \}_{n=1}^\infty$ is unbounded. Combining this with the second case solution yields $\lvert n^t \sin n \rvert \rightarrow +\infty$.

However, it's probably open problem which of $(A)$ or $(B)$ holds, along with the very value of $\mu$.

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  • $\begingroup$ Generalising this, to compute the irrationality measure of particular irrational number $\alpha$, we can instead investigate when $\lvert n^t \sin \frac{\pi}{\alpha}n \rvert$ tends to infinity, and vice versa. $\endgroup$ Mar 15 '16 at 3:04
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Consider the following two cases:

  1. $t<0$
  2. $t>0$

Note that $t=0$ reduces to $\lim_{n\rightarrow \infty}|\sin n|$.

For case 1, since $\lim_{n\rightarrow \infty}n^t=0$, and $\sin n$ is bounded by $-1\leq \sin n\leq 1$. It follows that $\lim_{n\rightarrow \infty}|n^t\sin n|=|0\cdot\lim_{n\rightarrow \infty}\sin n|=0$ (I'm abusing notation slightly here, technically you need to bound the absolute value with the 'worse case' (that being $\sin n=\pm 1$) and then use that bound to prove the limit).

For case 2, It's easy to find a subsequence that grows without bound, but I'm not sure how to get a bounded subsequence the converges to $0$ fast enough to prove that it doesn't diverge to infinity, but I strongly suspect that such a sequence exists. Edit: Nvm, see quartermind's answer (learn something new everyday).

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  • $\begingroup$ I don't see why, if $n_i$ is a sequence of integers with $|\sin n_i|<\epsilon$, it follows that $|n_i^t \sin n_i|$ is bounded, as would be necessary to prove that $|n^t\sin n|$ does not go to infinity as $n\to\infty$. It seems to me that no matter how small $\epsilon$ is, the sequence $|n_i^t \sin n_i|$ is merely shown to be $<\epsilon n_i^t$, and that is unbounded as $i\to\infty$. $\endgroup$
    – ForgotALot
    Mar 14 '16 at 2:42
  • $\begingroup$ @ForgotALot Good catch. $\endgroup$ Mar 14 '16 at 2:48
  • $\begingroup$ @JustinBenfield, your argument in case 2 is completely incorrect. $\endgroup$ Mar 14 '16 at 3:24
  • $\begingroup$ You revised your argument for case 2 since I wrote my comment, but it is still unconvincing. You don't just need a sequence such that $|\sin n_i|$ converges to zero; you need one that converges to zero fast enough to counteract the factor $n^t$ and keep the overall $|n_i^t\sin n_i|$ bounded. $\endgroup$
    – ForgotALot
    Mar 14 '16 at 4:13
  • $\begingroup$ I should've said converges to $0$ fast enough with the edit. $\endgroup$ Mar 14 '16 at 11:20

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