Answer
The radius of convergence of the series is $\dfrac{1}{4}$.
Work Step by Step
Here, we have $a_n=\dfrac{(4x)^n}{n^2}$
Ratio Test states that when $\Sigma a_n$ is an infinite series with positive terms and, then $r=|\lim\limits_{n \to \infty}\dfrac{a_{n+1}}{a_n}|$
a) When $0 \leq r \lt 1$, the series converges. (b) When $r \gt 1$, or, $\infty$, so the series diverges. (c) When $r=1$, the ratio test is inconclusive.
Now, $r=\lim\limits_{n \to \infty}|\dfrac{\dfrac{(4x)^{n+1}}{(n+1)^2}}{\dfrac{(4x)^n}{n^2}}|\\=\lim\limits_{n \to \infty}|\dfrac{4x(n^2)}{(n+1)^2}|\\=|4x|$
Therefore, the series converges for $|4x| \lt 1$ or, $|x| \lt \dfrac{1}{4}$ and diverges for $|4x| \gt 1$ by the ratio test. This implies that the radius of convergence of the series is $\dfrac{1}{4}$.
