Answer
series has radius of convergence is $\frac{1}{4}$
because the intervals of $x$ is from $-\frac{1}{4}$ to $\frac{1}{4}$ .
Work Step by Step
Root test:$R=\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=|\frac{\frac{(2n+2)!(x)^{n+1}}{((n+1)!)^{2}}}{\frac{2n!(x)^{n}}{(n!)^{2}}}|$
$=|\lim\limits_{n \to \infty}\frac{(2n+2)(2n+1}{(n+1)^{2}}.x|$
$=|\lim\limits_{n \to \infty}\frac{4n^{2}+6n+2}{n^{2}+2n+1}.x|$
$=|\frac{4.x}{1}|$
$=|4x|$
The series will converge when $|4x|\lt 1$ or $|x|\lt \frac{1}{4}$
Thus, the series has radius of convergence is $\frac{1}{4}$
because the intervals of $x$ is from $-\frac{1}{4}$ to $\frac{1}{4}$ .
