limit of $|n^t\sin n|$ 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 $t&gt;0$ (the reason is that the density implies that inequalities $\sin n&gt;\frac{1}{2}$ and $\sin n&lt;-\frac{1}{2}$ 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.
The question is strongly connected with irrationality measure of $\pi$. That is such number $\mu$, that for every numbers $\lambda, \nu$ with $\lambda &lt; \mu &lt; \nu$:


there exist infinitely many distinct rational numbers $p/q$ for which
$$
\left| \frac{p}{q} - \pi \right| &lt; \frac{1}{q^\lambda} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert &lt; q^{1-\lambda}
$$
for each rational $p/q$ with sufficiently large denominator
$$
\left| \frac{p}{q} - \pi \right| &gt; \frac{1}{q^\nu} \;\;\Longleftrightarrow\;\; \lvert p - \pi q \rvert &gt; q^{1-\nu}
$$


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 $t &gt; \mu-1$, then the limit is $+\infty$; if $t &lt; \mu-1$, then it doesn't exist.  The remaining case $t = \mu-1$ depends on the behavior of diophantine approximations of $\pi$.

First case: $t &lt; \mu-1$.

If $t &lt; \mu-1$ 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: $t &gt; \mu-1$.

Take $\varepsilon &gt; 0$, such that $t-\varepsilon &gt; \mu-1$. 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 $C &gt; 0$ 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 $\mu$.
Consider the following two cases:


$t&lt;0$
$t&gt;0$


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).