1. **Problem Statement:** Find the Taylor series for the function $f$ centered at $6$ given that $$f^{(n)}(6) = \frac{(-1)^n n!}{4^n (n+2)}$$ and determine the radius of convergence $R$ of this Taylor series. 2. **Taylor Series Formula:** The Taylor series of $f$ centered at $a=6$ is $$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(6)}{n!} (x-6)^n$$ 3. **Substitute the given derivative formula:** Substitute $f^{(n)}(6)$ into the series: $$f(x) = \sum_{n=0}^\infty \frac{(-1)^n n!}{4^n (n+2)} \cdot \frac{(x-6)^n}{n!} = \sum_{n=0}^\infty \frac{(-1)^n}{4^n (n+2)} (x-6)^n$$ 4. **Taylor series expression:** $$\boxed{f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{4^n (n+2)} (x-6)^n}$$ 5. **Radius of Convergence $R$:** Use the root test or ratio test. Consider the general term: $$a_n = \frac{(-1)^n}{4^n (n+2)} (x-6)^n$$ Focus on the absolute value: $$|a_n| = \frac{|x-6|^n}{4^n (n+2)}$$ Apply the root test: $$\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \frac{|x-6|}{4} \sqrt[n]{\frac{1}{n+2}} = \frac{|x-6|}{4} \cdot 1 = \frac{|x-6|}{4}$$ For convergence, this limit must be less than 1: $$\frac{|x-6|}{4} < 1 \implies |x-6| < 4$$ Thus, the radius of convergence is $$\boxed{R = 4}$$