1. **State the problem:** We are given the power series $$\sum_{n=1}^{\infty} \frac{(x-12)^n}{5^n \sqrt[3]{n}}$$ and need to find its radius of convergence $R$, and the interval endpoints $a$ and $b$ where the series converges. 2. **Formula for radius of convergence:** For a power series centered at $c=12$, $$\sum_{n=1}^{\infty} a_n (x-12)^n,$$ the radius of convergence $R$ is given by the Cauchy root test: $$R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}$$ where here $$a_n = \frac{1}{5^n \sqrt[3]{n}}.$$ 3. **Calculate $\limsup$:** Calculate $$\limsup_{n \to \infty} \sqrt[n]{\left|\frac{1}{5^n \sqrt[3]{n}}\right|} = \lim_{n \to \infty} \frac{1}{5 \sqrt[n]{\sqrt[3]{n}}} = \frac{1}{5} \lim_{n \to \infty} \frac{1}{n^{1/(3n)}}.$$ Since $$\lim_{n \to \infty} n^{1/(3n)} = 1,$$ we get $$\limsup = \frac{1}{5}.$$ 4. **Radius of convergence:** $$R = \frac{1}{\frac{1}{5}} = 5.$$ 5. **Interval of convergence:** The series converges for $$|x-12| < 5,$$ so $$7 < x < 17.$$ 6. **Check convergence at endpoints:** - At $x=a=7$, the series becomes $$\sum_{n=1}^{\infty} \frac{(7-12)^n}{5^n \sqrt[3]{n}} = \sum_{n=1}^{\infty} \frac{(-5)^n}{5^n \sqrt[3]{n}} = \sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt[3]{n}}.$$ This is an alternating series with terms decreasing to zero, so it converges by the Alternating Series Test. - At $x=b=17$, the series becomes $$\sum_{n=1}^{\infty} \frac{(17-12)^n}{5^n \sqrt[3]{n}} = \sum_{n=1}^{\infty} \frac{5^n}{5^n \sqrt[3]{n}} = \sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}.$$ This is a p-series with $p=\frac{1}{3} < 1$, which diverges. **Final answers:** $$R=5, \quad a=7, \quad b=17,$$ The series converges at $x=a=7$ and diverges at $x=b=17$.