Answer
The radius of convergence for the power series is $R=\infty$ and the interval of convergence is $(-\infty, \infty)$.
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 $a_k=\dfrac{(2x)^k}{k!}$
Now, $l=\lim\limits_{k \to \infty}|\dfrac{a_{k+1}}{a_k}|=\lim\limits_{k \to \infty}|\dfrac{\dfrac{(2x)^{k+1}}{(k+1)!}}{\dfrac{(2x)^k}{k!}}|=|2x| \lim\limits_{k \to \infty} \dfrac{1}{k+1}=0$
So, the radius of convergence for the power series is $R=\dfrac{1}{l}=\dfrac{1}{0}=\infty$. This implies that the interval of convergence is $(-\infty, \infty)$.
