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