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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 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.

Please help me how to 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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