Given number:
$$\frac{\sin^{n}{\left(3 \right)}}{3^{n}}$$
It is a series of species
$$a_{n} \left(c x - x_{0}\right)^{d n}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{n \to \infty} \left|{\frac{a_{n}}{a_{n + 1}}}\right|}{c}$$
In this case
$$a_{n} = \sin^{n}{\left(3 \right)}$$
and
$$x_{0} = -3$$
,
$$d = -1$$
,
$$c = 0$$
then
$$\frac{1}{R} = \tilde{\infty} \left(-3 + \lim_{n \to \infty}\left(\sin^{n}{\left(3 \right)} \sin^{- n - 1}{\left(3 \right)}\right)\right)$$
Let's take the limitwe find
False
False
$$R = 0$$