Sum of series sin2n

Given number:
$$\sin{\left(2 n \right)}$$
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
$$a_{n} = \sin{\left(2 n \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n \right)}}{\sin{\left(2 n + 2 \right)}}}\right|$$
Let's take the limit
we find
$$1 = \lim_{n \to \infty} \left|{\frac{\sin{\left(2 n \right)}}{\sin{\left(2 n + 2 \right)}}}\right|$$

False