Given number:
$$\sqrt{\frac{k + 1}{k}} \log{\left(k \right)}$$
It is a series of species
$$a_{k} \left(c x - x_{0}\right)^{d k}$$
- power series.
The radius of convergence of a power series can be calculated by the formula:
$$R^{d} = \frac{x_{0} + \lim_{k \to \infty} \left|{\frac{a_{k}}{a_{k + 1}}}\right|}{c}$$
In this case
$$a_{k} = \sqrt{\frac{k + 1}{k}} \log{\left(k \right)}$$
and
$$x_{0} = 0$$
,
$$d = 0$$
,
$$c = 1$$
then
$$1 = \lim_{k \to \infty}\left(\frac{\left(k + 1\right) \left|{\log{\left(k \right)}}\right|}{\sqrt{k} \sqrt{k + 2} \log{\left(k + 1 \right)}}\right)$$
Let's take the limitwe find
True
False