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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?

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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 $1$.

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Hint: what does $x^n$ do as $n \to \infty$ if $|x| < 1$? If $|x| > 1$?

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