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

But I found the solution by trial and error.

Substituting $$1$$ for $$x$$, the series diverges, so $$2$$ and $$4$$ cannot be the radii of convergence.

Substituting $$1/2$$, the series converges. So the answer is $$1$$.

How does one usually proceed in problem like this?

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 $$|x| > 1$$?