Answer
The radius of convergence for the power series is $R=0$ and the interval of convergence is $\{x | x=0\}$.
Work Step by Step
Root Test; Let us consider an infinite series $\Sigma a_k$ with positive terms and suppose that $l=\lim\limits_{k \to \infty} \sqrt[k] a_k$
1) When $0 \leq l \lt 1$. then the series $\Sigma a_k$ converges
2) When $l \gt 0$. 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 $a_k=(kx)^k$
Now, $l=\lim\limits_{k \to \infty} \sqrt[k] {(kx)^{k}}=|x| \lim\limits_{k \to \infty} (1+\dfrac{1}{k})^k$
This implies that that for every $x \ne 0$ we have $l=\infty$, so the power series diverges So, the radius of convergence for the power series is $R=0$ and the interval of convergence is $\{x | x=0\}$.
