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Can someone verify this?

Suppose that the series $$\sum\limits_{n=1}^\infty a_n x^n$$ has a radius of convergence $R$, where $0 < R < \infty$

(a) Find the radius of convergence of $\sum\limits_{n=1}^\infty a_nx^{2n}$

(b) Find the radius of convergence of $\sum\limits_{n=1}^\infty a_n^2x^n$

Note that

$$ \operatorname{lim sup} |a_n|^{\frac{1}{n}} = \frac{1}{R}$$

(a) The series converges if $$\operatorname{lim sup } |a_nx^{2n}|^{\frac{1}{n}} < 1$$

That is, $$x^2\operatorname{lim sup } |a_n|^{\frac{1}{n}} < 1$$ $$x^2 < \frac{1}{R}$$ $$|x| < \sqrt{\frac{1}{R}}$$

(b) The series converges if $$\operatorname{lim sup} |a_n^2x^n|^{\frac{1}{n}} < 1$$

That is, $$|x|\operatorname{lim sup} \left(|a_n|^{\frac{1}{n}}\right)^2 < 1$$

$$|x| < \frac{1}{R^2}$$

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From what you have you know that
$$\limsup |a_n|^{\frac{1}{n}}=\frac{1}{R}\ .$$ For tha case (a) you know that $$\frac{1}{R_a}=\limsup |a_n|^{\frac{1}{2n}}=\frac{1}{\sqrt{R}}\ .$$ In this case the sequence you are considering in the $\limsup$ is basically $\{0,a_1,0,a_2,0,a_3,0, a_4,\dots\}$, whose $\limsup$ is the same of $\{a_1,a_2,a_3,a_4,\dots\}$.

For tha case (b) you have that $$\frac{1}{R_b}=\limsup |a^2_n|^{\frac{1}{n}}=(\limsup |a_n|^{\frac{1}{n}})^2=\frac{1}{R^2}\ .$$

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