## University Calculus: Early Transcendentals (3rd Edition)

Radius of convergence is $3$
We are given that:$a_n=\dfrac{n!}{3 \cdot 6.....3n}x^n$ We need to apply the Ratio Test to the series. $\lim\limits_{n \to \infty} |\dfrac{a_{n+1}}{a_n}|=\lim\limits_{n \to \infty} \dfrac{a_n (x/3)}{a_n}$ or, $=\lim\limits_{n \to \infty} |\dfrac{x}{3}|$ Now, $|\dfrac{x}{3}| \lt 1 \implies |x| \lt 3$ Thus, the radius of convergence is $3$.