As Robin Chapman mentions in his comment, the difficulty of investigating
the convergence of
$$
\sum_{n=1}^\infty\frac1{n^3\sin^2n}
$$
is due to lack of knowledge about the behavior of $|n\sin n|$ as $n\to\infty$,
while the latter is related to rational approximations to $\pi$ as follows.

Neglecting the terms of the sum for which $n|\sin n|\ge n^\varepsilon$
($\varepsilon>0$ is arbitrary),
as they all contribute only to the `convergent part' of the sum, the question
is equivalent to the one for the series
$$
\sum_{n:n|\sin n|< n^\varepsilon}\frac1{n^3\sin^2n}.
\qquad(1)
$$
For any such $n$, let $q=q(n)$ minimizes the distance $|\pi q-n|\le\pi/2$.
Then
$$
\sin|\pi q-n|=|\sin n|< \frac1{n^{1-\varepsilon}},
$$
so that $|\pi q-n|\le C_1/n^{1-\varepsilon}$ for some absolute constant $C_1$
(here we use that $\sin x\sim x$ as $x\to0$). Therefore,
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_1}{qn^{-\varepsilon}},
$$
equivalently
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_2}{n^{2-\varepsilon}}
\quad\text{or}\quad
\biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}}
$$
(because $n/q\approx\pi$) for all $n$ participating in the sum (1).
It is now clear that the convergence of the sum (1) depends
on how often we have
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_2'}{q^{2-\varepsilon}}
$$
and how small is the quantity in these cases. (Note that
it follows from Dirichlet's theorem that an even stronger inequality,
$$
\biggl|\pi-\frac nq\biggr|<\frac1{q^2},
$$
happens for infinitely many pairs $n$ and $q$.)
The series (1) converges if and only if
$$
\sum_{n:|\pi-n/q|< C_2n^{-2+\varepsilon}}\frac1{n^5|\pi-n/q|^2}
$$
converges. We can replace the summation by summing over $q$
(again, for each term $\pi q\approx n$) and then sum the result
over all $q$, because the
terms corresponding to $|\pi-n/q|< C_2n^{-2+\varepsilon}$ do not
influence on the convergence:
$$
\sum_{q=1}^\infty\frac1{q^5|\pi-n/q|^2}
=\sum_{q=1}^\infty\frac1{q^3(\pi q-n)^2}
\qquad(2)
$$
where $n=n(q)$ is now chosen to minimize $|\pi-n/q|$.

Summarizing, *the original series converges if and only if the series in* (2)
*converges.*

It is already an interesting question of what can be said about the
convergence of (2) if we replace $\pi$ by other constant $\alpha$,
for example by a "generic irrationality". The series
$$
\sum_{q=1}^\infty\frac1{q^3(\alpha q-n)^2}
$$
for a real quadratic irrationality $\alpha$ converges because the best
approximations are $C_3/q^2\le|\alpha-n/q|\le C_4/q^2$, and they are achieved on
the convergents $n/q$ with $q$ increasing geometrically. A more delicate
question seems to be for $\alpha=e$, because one third of its convergents satisfies
$$
C_3\frac{\log\log q}{q^2\log q}<\biggl|e-\frac pq\biggr|< C_4\frac{\log\log q}{q^2\log q}
$$
(see, e.g., [C.S.Davis, *Bull. Austral. Math. Soc.* 20 (1979) 407--410]).
The number $e$, quadratic irrationalities, and even algebraic numbers
are `generic' in the sense that their irrationality exponent is known to be 2.
What about $\pi$?

The *irrationality exponent* $\mu=\mu(\alpha)$ of a real irrational number
$\alpha$ is defined as the infimum of exponents $\gamma$
such that the inequality $|\alpha-n/q|\le|q|^{-\gamma}$ has
only finitely many solutions in $(n,q)\in\Bbb Z^2$ with $q\ne0$.
(So, Dirichlet's theorem implies that $\mu(\alpha)\ge2$. At the same
time from metric number theory we know that it is 2 for almost all real irrationals.)
Assume that $\mu(\pi)>5/2$, then there are infinitely many
solutions to the inequality
$$
\biggl|\pi-\frac nq\biggr|<\frac{C_5}{q^{5/2}},
$$
hence infinitely many terms in (2) are bounded below by $1/C_5$, so that
the series diverges (and (1) does as well). Although the general
belief is that $\mu(\pi)=2$, the best known result of V.Salikhov (see
this answer
by Gerry and my comment)
only asserts that $\mu(\pi)<7.6064\dots$,.

I hope that this explains the problem of determining the behavior of the series in question.