Find radius of convergence of the power series.

Find the radius of convergence of power series $$\sum_{n=0}^{\infty} 2^{2n} x^{n^2}$$ A)1

B)2

C) 4

D)1/4

I try to apply ratio and root test ( Cauchy–Hadamard theorem ) .but they don't seem promising as I was left with $$|x^n|$$ .

But I find solution by trial and error . I substitute 1 for x , series diverges , so 2,4 cannot be radius of convergence, and I substitute 1/2 the series converges .so the answer is 1.

Since$$\left\lvert\frac{2^{2(n+1)}x^{(n+1)^2}}{2^{2n}x^{n^2}}\right\rvert=2^2\lvert x\rvert^{2n+1},$$the series converges absolutely if $$\lvert x\rvert<1$$ and diverges if $$\lvert x\rvert>1$$. So, the radius of convergence is $$1$$.
Hint: what does $$x^n$$ do as $$n \to \infty$$ if $$|x| < 1$$? If $$|x| > 1$$?