1. **State the problem:** Find the radius and interval of convergence for the series $$\sum_{n=1}^\infty \frac{2n! (4x - 5)^n}{17 n^3}$$ with center at $$x = \frac{5}{4}$$. 2. **Recall the formula for radius of convergence:** For a power series $$\sum a_n (x - c)^n$$, the radius of convergence $$R$$ is given by $$ \frac{1}{R} = \limsup_{n \to \infty} |a_n|^{1/n} $$ where $$a_n$$ are the coefficients of the series. 3. **Identify the coefficients:** Here, $$a_n = \frac{2 n!}{17 n^3}$$ and the series is in terms of $$4x - 5$$, so the center is $$c = \frac{5}{4}$$. 4. **Apply the root test to find radius:** Calculate $$ \limsup_{n \to \infty} \left| \frac{2 n!}{17 n^3} \right|^{1/n} = \lim_{n \to \infty} \frac{2^{1/n} (n!)^{1/n}}{17^{1/n} n^{3/n}} = \lim_{n \to \infty} (n!)^{1/n} $$ (since $$2^{1/n} \to 1$$, $$17^{1/n} \to 1$$, and $$n^{3/n} \to 1$$). 5. **Use Stirling's approximation:** $$ n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n $$ Taking the $$n$$th root, $$ (n!)^{1/n} \approx \left( \sqrt{2 \pi n} \right)^{1/n} \frac{n}{e} \to \frac{n}{e} \text{ as } n \to \infty $$ 6. **Evaluate the limit:** $$ \lim_{n \to \infty} (n!)^{1/n} = \lim_{n \to \infty} \frac{n}{e} = \infty $$ 7. **Calculate radius:** $$ \frac{1}{R} = \infty \implies R = 0 $$ 8. **Interpretation:** Radius of convergence is zero, so the series converges only at the center $$x = \frac{5}{4}$$. 9. **Interval of convergence:** Since $$R=0$$, the interval of convergence is just the single point $$ \boxed{x = \frac{5}{4}} $$ **Final answer:** - Radius of convergence $$R = 0$$ - Interval of convergence $$\{ x = \frac{5}{4} \}$$