Answer
The radius of convergence: $3$
The interval of convergence: $[-3,3)$
Work Step by Step
Let $a_n=\frac{x^n}{n3^n}$.
It can be checked that $\frac{a_{n+1}}{a_n}=\frac{(n+1)x}{3n}$.
Then,
$\lim |\frac{a_{n+1}}{a_n}|=\lim|\frac{(n+1)x}{3n}|=|\frac{x}{3}|$.
Find the interval of convergence:
$|\frac{x}{3}|<1$
$|x|<3$
For $x=-3$, the series would be $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ and it is convergent.
For $x=3$, the series would be a harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ and it is divergent.
Thus, the radius of convergence is $3$ and the interval of convergence is $[-3,3)$.
