$$\sum_{n=1}^\infty \frac{2^nx^{2n}}{n^2}$$
Hello, I am having trouble knowing where to go with problem. I need to find the interval I on which this converges uniformly. I suspect the interval is at least |x| < 1/2 since I know and found when x = 1/2 it cancels the $2^n$ allowing for uniform convergence. However, I am not sure how to go forward and find a possibly bigger interval for this and would appreciate help.
Let $$u_n(x)=\frac{2^nx^{2n}}{n^2}$$ For $ x\ne 0$,
the ratio test gives $$\lim_{n\to+\infty}\frac{u_{n+1}(x)}{u_n(x)}=2x^2$$
thus
$$x^2<\frac 12 \implies \sum u_n(x)\text { converges}$$
$$x^2>\frac 12\implies \sum u_n(x)\text{ diverges}$$
and $$x^2=\frac 12\implies u_n(x)=\frac{1}{n^2}\implies \sum u_n(x)\text{ converges}$$