Sum of series (5n-8)^(4n+1)/(8n)

Given number:
$$\frac{\left(5 n - 8\right)^{4 n + 1}}{8 n}$$
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
$$a_{n} = \frac{\left(5 n - 8\right)^{4 n + 1}}{8 n}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(n + 1\right) \left|{\left(5 n - 8\right)^{4 n + 1}}\right|}{n \left|{\left(5 n - 3\right)^{4 n + 5}}\right|}\right)$$
Let's take the limit
we find

False

False