Answer
The radius of convergence is $R=\sqrt 3$ and the interval of convergence is $(-\sqrt 3, \sqrt 3)$ .
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 $a_k=\dfrac{x^{2k+1}}{3^{k-1}}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{\dfrac{x^{2(k+1)+1}}{3^{(k+1)-1}}}{\dfrac{x^{2k+1}}{3^{k-1}}}| =\lim\limits_{k \to \infty} \dfrac{x^2}{3}=\dfrac{x^2}{3}$
This means that the series converges for $r=\dfrac{x^2}{3} \lt 1 \implies -\sqrt 3 \lt x \lt \sqrt 3$ and so, the radius of convergence for the power series which is centered at $0$ is $R=\sqrt 3$. Next, we will compute the interval of convergence for the two points. Those points are $-\sqrt 3$ and $\sqrt 3$.
Case 1: For $x=-\sqrt 3$
$\sum_{k=1}^{\infty} \dfrac{(-\sqrt 3)^{2k+1}}{3^{k-1}}=- \sum_{k=1}^{\infty} 3^{3/2}$; this implies that the series diverges.
Case 2:For $x=\sqrt 3$
$\sum_{k=1}^{\infty} \dfrac{(\sqrt 3)^{2k+1}}{3^{k-1}}= \sum_{k=1}^{\infty} 3^{3/2}$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=\sqrt 3$ and the interval of convergence is $(-\sqrt 3, \sqrt 3)$ .
