If $∑ a_nz ^n$ has radius of convergence R, use the formula $R = \left( \limsup_{n \to \infty} | a_n |  ^{\frac{1}{n}} \right)^{-1} $ to find the radius of convergence of $∑ a_nz^{ 3n}$ I tried to solve this question using the ratio test, I got an answer of $R^{\frac{1}{3}}$, however, I am not sure how to solve it using Cauchy-Hadamard formula. Any hint will be appreciated.
Setting: Let $\sum a_n z^n$ have radius of convergence $R$.  We have that  $$ R = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} $$  via Hadamard's formula for the radius of convergence.  Question: What are the radii of convergence for (i) $\sum a_n z^{2n}$ and (ii) $\sum a_n^2 z^n$?  Attempt for $\sum a_n^2 z^n$:   For ease of notation, let $R_1$ and $R_2$ denote the radii of convergence for power series $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ respectively. From Hadamard's formula, we then have  $$ R_2 = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n^2 \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} \right)^2 = R^2 $$   But what about $R_1$?
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