Answer
The radius of convergence is $R=3$ and the interval of convergence is $(-3,3)$ .
Work Step by Step
Ratio Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|$
1) When $l \lt 1$. then series $\Sigma a_k$ converges absolutely.
2) When $l \gt 1$. then series $\Sigma a_k$ diverges.
3) When $l = 1$. then series $\Sigma a_k$ may converge or diverge, that is, the test is inconclusive.
Here, we have $a_k=(\dfrac{x}{3})^k$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{(\dfrac{x}{3})^{k+1}}{(\dfrac{x}{3})^k}|= \lim\limits_{k \to \infty} |\dfrac{x}{3}|=|\dfrac{x}{3}|$
This means that the series converges for $r=|\dfrac{x}{3}| \lt 1$ . Then $-1 \lt \dfrac{x}{3} \lt 1 \implies -3 \lt x \lt 3$.
So, the radius of convergence for the power series is $R=3$
Next, we will compute the interval of convergence for the two points. Those points are $-3$ and $3$.
Case 1: For $x=-3$
$\sum_{k=1}^{\infty} (\dfrac{-3}{3})^k=\sum_{k=1}^{\infty} (-1)^k$; this implies that the series diverges.
Case 2: For $x=3$
$\sum_{k=1}^{\infty} (\dfrac{-3}{3})^k=\sum_{k=1}^{\infty} (1)^k$; this implies that the series converges.
Thus, we conclude that the radius of convergence is $R=3$ and the interval of convergence is $(-3,3)$ .
