Answer
The radius of convergence is $R=5$ and the interval of convergence is $(-5,5)$ .
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 $a_k=(-1)^k \dfrac{x^k}{5^k}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{(-1)^{k+1} \dfrac{x^{k+1}}{5^{k+1}}}{(-1)^k \dfrac{x^k}{5^k}}|= \lim\limits_{k \to \infty} |\dfrac{x}{5}|=|\dfrac{x}{5}|$
This means that the series converges for $r=|\dfrac{x}{5}| \lt 1$ . Then $-1 \lt \dfrac{x}{5} \lt 1 \implies -5 \lt x \lt 5$.
So, the radius of convergence for the power series is $R=5$
Next, we will compute the interval of convergence for the two points. Those points are $-5$ and $5$.
Case 1: For $x=-5$
$\sum_{k=1}^{\infty} (-1)^k \dfrac{(-5)^k}{5^k}=\sum_{k=1}^{\infty} 1=\infty$; this implies that the series diverges.
Case 2: For $x=5$
$\sum_{k=1}^{\infty} (-1)^k \dfrac{(5)^k}{5^k}=\sum_{k=1}^{\infty} (-1)^k$; this implies that the series diverges.
Thus, we conclude that the radius of convergence is $R=5$ and the interval of convergence is $(-5,5)$ .
