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