Show that $(\sum a_{n}^{3} \sin n)$ converges given $\sum{a_n}$ converges Given that $\sum a_{n}$ converges $\left(a_{n}>0\right) ;$ Then $(\sum a_{n}^{3} \sin n)$ is

My approach:
Since, $\sum a_{n}$ converges, we have $\lim _{n \rightarrow \infty} n \cdot a_{n}$ converges.
i.e. $\left|n \cdot a_{n}\right| \leq 1$ for $n \geq K(\text { say })$
$\Rightarrow n \cdot a_{n}<1 \quad\left[\because a_{n}>0\right]$
$\Rightarrow a_{n}<\frac{1}{n}$
$\therefore a_{n}^{3}<\frac{1}{n^{3}}$
$\Rightarrow a_{n}^{3} \sin n \leq \frac{1}{n^{3}} \sin n \leq \frac{1}{n^{3}}$
$\Rightarrow \sum a_{n}^{3} \sin n \leq \sum \frac{1}{n^{3}}$
$\because \mathrm{RHS}$ converges so LHS will also converge.
Any other better approach will be highly appreciated and correct me If I am wrong
 A: As $\{a_n\}$ is a positive sequence such that $\sum a_n$ converges, we have $0 \le a_n \le 1$ for $n$ large enough, say $n \ge M$.
Then for $n \ge M$
$$0 \le \vert a_n^3 \sin n \vert \le a_n^3 \le a_n$$
Hence $\sum a_n^3 \sin n$ converges absolutely.
Also a series $\sum a_n$ can be convergent while the sequence $\{n a_n\}$ diverges. Consider for example $a_n$ equals to $0$ if $n$ is not a square and equal to $1/n$ otherwise.
A: For $a_n\geq 0$ the sum converges as shown by @mathcounterexamaples.net
For general $a_n$ the statement is not true. The (deleted) counter-example given by @Mark Viola indeed is a counter-example. Taking $a_n=\sin(n)/n^{1/3}$ it suffices to show that $\sum_{n\geq 1} a_n^3 \sin(n) =\sum_{n\geq 1} \frac{\sin^4 n}{n} =+\infty$. To see this note that $n$ is asymptotically equidistributed mod $2\pi$ which implies that for any continuous function $f\in C([-1,1])$, $ \lim_{N\rightarrow +\infty} \frac{1}{N} \sum_{n=1}^N f(\sin(n)) = 
   \frac{1}{2\pi} \int_0^{2\pi} f(\sin(t))\; dt$. In particular, as $N\rightarrow +\infty$:
$$ M_N =  \frac{1}{N} \sum_{n=1}^N \sin^4(n) \rightarrow 
   \frac{1}{2\pi} \int_0^{2\pi} \sin^4(t)\; dt = \frac{3}{8}.$$
Proceeding as in Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$   find a strictly increasing sequence $N_k$, $k\geq 1$ so that each $M_{N_k}\geq \frac{1}{4}$ and $8 N_k \leq N_{k+1}$. Then
$$ \sum_{n=1+N_k}^{N_{k+1}} \frac{\sin^4(n)}{n} \geq \frac{1}{N_{k+1}} \sum_{n=1+N_k}^{N_{k+1}} \sin^4(n) \geq M_{N_{k+1}} -\frac{N_k}{N_{k+1}} \geq \frac{1}{4} - \frac{1}{8}= \frac{1}{8} $$
Summing over $k$ implies the divergence of $\sum_{n\geq 1} \frac{\sin^4(n)}{n}$.
