Chapter 9.8, Problem 3QCE

Since $\underset{k\to +\infty }{\lim }\left[\frac{\left(3^{k+1}{x}^{k+1}\right)/\left(k+1\right)!}{(3^k{x}^k)/k!}\right]=\underset{k\to +\infty }{\lim }\left|\frac{3x}{k+1}\right|=0$ the interval of convergence for the series ${\displaystyle {\sum }_{k=0}^{\infty }(3^k/k!)x^k}\text{is}\text{\_\_\_\_\_\_\_\_}\text{.}$