I tried to use the ratio test to determine the convergence interval for the series of functions given as $\Sigma^{\infty}_{k=0}\frac{k!(x-2)^k}{k^k}$. But the ratio test was infinity no matter the x. Here is the working:
$$\Sigma^{\infty}_{k=0}\frac{k!(x-2)^k}{k^k} \\ \text{ Izmanto Dalambēra testu } \frac{a_{n+1}}{a_n} = \frac{(k+1)!(x-2)^{k+1}k^k}{k^{k+1}k!(x-2)^k} = \frac{(x-2)^k (k+1)}{k} \\ \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k} \right| = \infty$$
Maybe there is a better way to find the convergence interval?
Your computations are not correct. We have\begin{align}\frac{\left|\frac{(k+1)!(x-2)^{k+1}}{(k+1)^{k+1}}\right|}{\left|\frac{k!(x-2)^k}{k^k}\right|}&=\left(\frac k{k+1}\right)^k|x-2|\\&\to\frac{|x-2|}e.\end{align}Therefore, the radius of convergence is $e$.