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.
