Show that $(sum a_{n}^{3} sin n)$ converges given $sum{a_n}$ converges

For $a_ngeq 0$ 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 $sum_{ngeq 1} a_n^3 sin(n) =sum_{ngeq 1} frac{sin^4 n}{n} =+infty$. To see this note that $n$ is asymptotically equidistributed mod $2pi$ which implies that for any continuous function $fin C([-1,1])$, $ lim_{Nrightarrow +infty} frac{1}{N} sum_{n=1}^N f(sin(n)) = frac{1}{2pi} int_0^{2pi} f(sin(t)); dt$. In particular, as $Nrightarrow +infty$: $$ M_N = frac{1}{N} sum_{n=1}^N sin^4(n) rightarrow frac{1}{2pi} int_0^{2pi} sin^4(t); dt = frac{3}{8}.$$ Proceeding as in Convergence of $sum_n frac{|sin(n^2)|}{n}$ find a strictly increasing sequence $N_k$, $kgeq 1$ 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 $sum_{ngeq 1} frac{sin^4(n)}{n}$.

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.