Given number:
$$\frac{\sqrt{n + 1}}{\left(n^{5} + 2 n\right) + 1}$$
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
$$a_{n} = \frac{\sqrt{n + 1}}{n^{5} + 2 n + 1}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\sqrt{n + 1} \left(2 n + \left(n + 1\right)^{5} + 3\right)}{\sqrt{n + 2} \left(n^{5} + 2 n + 1\right)}\right)$$
Let's take the limitwe find
True
False