Answer
The radius of convergence is $R=10$ and the interval of convergence is $(-10, 10)$ .
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 $a_k=(-\dfrac{x}{10})^{2k}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{(-\dfrac{x}{10})^{2(k+1)}}{(-\dfrac{x}{10})^{2k}}| =\lim\limits_{k \to \infty} |(\dfrac{x}{10})^2|=\dfrac{x^2}{100}$
This means that the series converges for $r=\dfrac{x^2}{100} \lt 1 \implies -10 \lt x \lt 10$ and so, the radius of convergence for the power series which is centered at $0$ is $R=10$. Next, we will compute the interval of convergence for the two points. Those points are $-10$ and $10$.
Case 1: For $x=-10$
$\sum_{k=1}^{\infty} (-\dfrac{-10}{10})^{2k}= \sum_{k=1}^{\infty} 1$; this implies that the series diverges.
Case 2:For $x=10$
$\sum_{k=1}^{\infty} (-\dfrac{10}{10})^{2k}= \sum_{k=1}^{\infty} 1=\infty$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=10$ and the interval of convergence is $(-10, 10)$ .
