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