Given number:
$$n \sin^{2}{\left(\frac{\pi}{n^{3}} \right)}$$
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
$$a_{n} = n \sin^{2}{\left(\frac{\pi}{n^{3}} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{n \sin^{2}{\left(\frac{\pi}{n^{3}} \right)} \left|{\frac{1}{\sin^{2}{\left(\frac{\pi}{\left(n + 1\right)^{3}} \right)}}}\right|}{n + 1}\right)$$
Let's take the limitwe find
True
False