Given number:
$$\frac{1}{\left(n^{2} + 13 n\right) + 40}$$
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
$$a_{n} = \frac{1}{n^{2} + 13 n + 40}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{13 n + \left(n + 1\right)^{2} + 53}{n^{2} + 13 n + 40}\right)$$
Let's take the limitwe find
True
False