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