Answer
$R=1$ ; interval of convergence is $[-1,1]$
Work Step by Step
Let $a_{n}=\frac{x^{2n}}{n(lnn)^{2}}$, then
$\lim\limits_{n \to \infty}|\frac{a_{n+1}}{a_{n}}|=\lim\limits_{n \to \infty}|\dfrac{\frac{x^{2(n+1)}}{(n+1)(ln(n+1))^{2}}}{\frac{x^{2n}}{n(lnn)^{2}}}|$
$=x^{2}\lt 1$
Thus, $-1\lt x\lt 1$
Hence, $R=1$ ; interval of convergence is $[-1,1]$
