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