91Ó°ÊÓ

First, let's apply the Ratio Test to the given power series. The Ratio Test states that the radius of convergence is given by: $$R = \lim_{k \to \infty} \left|\frac{a_k}{a_{k+1}}\right|$$ where \(a_k\) is the general term of the power series. For the given power series, we have \(a_k = k!(x-10)^k\). Then, the next term is \(a_{k+1} = (k+1)!(x-10)^{k+1}\).

Now, let's compute the limit involved in the Ratio Test. $$R = \lim_{k\to\infty} \left|\frac{a_k}{a_{k+1}} \right| = \lim_{k\to\infty} \left|\frac{k!(x-10)^k}{(k+1)!(x-10)^{k+1}}\right|$$ To simplify, we can cancel out \((x-10)^k\) from both the numerator and denominator and expand (k+1)! $$R = \lim_{k\to\infty} \left|\frac{k!}{(k+1)k!}\right|\cdot\left|\frac{1}{x-10}\right|$$ The k! terms cancel each other, and we are left with: $$R=\lim_{k\to\infty} \left|\frac{1}{k+1}\right|\cdot\left|\frac{1}{x-10}\right|$$ Since the limit of \(\frac{1}{k + 1}\) as \(k \to \infty\) is 0, the radius of convergence R is given by: $$R = \frac{1}{|x - 10|} \cdot 0 = 0$$ But this is not possible, the radius must be positive. In this case, the Ratio Test is inconclusive and we must try a different method.

Let's try using the Root Test instead. The Root Test states that the radius of convergence is also given by: $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|a_k|}}$$ Therefore, we have: $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|k!(x-10)^k|}}$$

Let's compute the limit involved in the Root Test. $$R = \frac{1}{\lim_{k \to \infty} \sqrt[k]{|k!(x-10)^k|}} = \frac{1}{\lim_{k \to \infty} |(x-10)\sqrt[k]{k!}|}$$ Now, the original power series converges if the limit of the expression inside the absolute value is less than 1. In other words, we are looking for the values of x for which $$|(x-10)\sqrt[k]{k!}| < 1$$ As \(k \to \infty\), according to the Stirling's approximation formula, \(\sqrt[k]{k!} \sim \sqrt{2\pi k}\left(\frac{k}{e}\right)^k\). Therefore, $$R = \frac{1}{\lim_{k \to \infty} |(x-10) \sqrt{2\pi k}\left(\frac{k}{e}\right)^k|}$$ Since Stirling's approximation grows rapidly with k, the only way this limit can be other than 0 is if \((x-10) = \infty\). Thus, the power series diverges for all x.

Since the power series diverges for all x, there is no need to test the endpoints and the interval of convergence is an empty set.