Given number:
$$\left(2 n + 6\right) \tan{\left(\frac{15}{9 n + 11} \right)}$$
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
$$a_{n} = \left(2 n + 6\right) \tan{\left(\frac{15}{9 n + 11} \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{n \to \infty}\left(\frac{\left(2 n + 6\right) \left|{\frac{\tan{\left(\frac{15}{9 n + 11} \right)}}{\tan{\left(\frac{15}{9 n + 20} \right)}}}\right|}{2 n + 8}\right)$$
Let's take the limitwe find
True
False