Finding Radii of Convergence for $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ 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 $R_1$?

Your calculation for the second is correct. For the first one note that
$$\sum_{n=0}^\infty a_n z^{2n} = \sum_{n=0}^\infty b_n z^n$$ 
with $b_n=a_{k}$ if $n=2k$ for $k\in\mathbb{N}$ and $b_n=0$ otherwise.
Then
$$R_1=\frac{1}{\limsup_{n\rightarrow\infty}|b_n|^{1/n}}=\frac{1}{\limsup_{n\rightarrow\infty} |a_n|^{1/(2n)}}=\left(\frac{1}{\limsup_{n\rightarrow\infty} |a_n|^{1/n}}\right)^{1/2}=R^{1/2}$$
