Does this sum converge or diverge?

$$ sum limits_{n=0}^{infty}frac{sin(n)cdot(n^2+3)}{2^n} $$

To solve this I would use $$ sin(z) = sum limits_{n=0}^{infty}(-1)^nfrac{^{2n+1}}{(2n+1)!} $$

and make it to $$sum limits_{n=0}^{infty}sin(n)cdotfrac{(n^2+3)}{2^n} = sum limits_{n=0}^{infty}(-1)^nfrac{n^{2n+1}}{(2n+1)!} cdot sum limits_{n=0}^{infty}frac{(n^2+3)}{2^n} $$

and since $$sum limits_{n=0}^{infty}frac{(n^2+3)}{2^n} text{ and } sum limits_{n=0}^{infty}(-1)^nfrac{n^{2n+1}}{(2n+1)!} $$

converges $$ sum limits_{n=0}^{infty}frac{sin(n)cdot(n^2+3)}{2^n} $$

would also converge.

Is my assumption true? I’m also a bit scared to use it since I’ve got the sin(z) equation from a source outside the stuff that my professor gave us