Answer
The interval of convergence for the series is: $(\dfrac{-1}{2}, \dfrac{1}{2}]$ and the radius of convergence is $R=\dfrac{1}{2}$.
Work Step by Step
Ratio Test: Let us consider a series $\Sigma a_k$ whose limit $l$ can be obtained as: $ l=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|$
1. For $l \lt 1$, the series is absolutely convergent.
2. For $l \gt 1$, the series is divergent.
3. For $l = 1$, the series is inconclusive.
Therefore, $ L=\lim\limits_{k \to \infty} |\dfrac{a_{k+1}}{a_k}|\\=\lim\limits_{k \to \infty} \dfrac{2^{k+1}x^{k+2}}{k+2} \times \dfrac{k+1}{2^k x^{k+1}} \\=\lim\limits_{k \to \infty} \dfrac{2(k+1)}{k+2}|x|\\=2 |x|$
So, we can conclude that the given series converges absolutely if $l=2|x| \lt 1$ for $-1\lt 2x \lt 1$ or $\dfrac{-1}{2} \lt x \lt \dfrac{1}{2}$. It diverges if $l=2|x| \gt 1$ .
Therefore, the interval of convergence for the series is: $(\dfrac{-1}{2}, \dfrac{1}{2}]$ and the radius of convergence is $R=\dfrac{1}{2}$.
