Radius of convergence of power series $\sum c_n x^{2n}$ and $\sum c_n x^{n^2}$ I've got a start on the question I've written below. I'm hoping for some help to finish it off.

Suppose that the power series $\sum_{n=0}^{\infty}c_n x^n$ has a radius of convergence $R \in (0, \infty)$. Find the radii of convergence of the power series $\sum_{n=0}^{\infty}c_n x^{2n}$ and $\sum_{n=0}^{\infty}c_n x^{n^2}$.

From Hadamard's Theorem I know that the radius of convergence for $\sum_{n=0}^{\infty}c_n x^n$ is $R=\frac{1}{\alpha}$, where
$$\alpha = \limsup_{n \to \infty} |a_n|^{\frac{1}{n}}.$$
Now, applying the Root Test to $\sum_{n=0}^{\infty}c_n x^{2n}$ gives
$$\limsup |a_nx^{2n}|^{\frac{1}{n}}=x^2 \cdot \limsup |a_n|^{\frac{1}{n}}=x^2 \alpha$$
which gives a radius of convergence $R_1 = \frac{1}{\sqrt{\alpha}}$. Now for the second power series. My first thought was to take
$$\limsup |a_nx^{n^2}|^{\frac{1}{n^2}}=|x| \cdot \limsup |a_n|^{\frac{1}{n^2}}$$
but then I'm stuck. I was trying to write the radius of convergence once again in terms of $\alpha$. Any input appreciated and thanks a bunch.

$$
\limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n}}=\alpha <\infty
$$
gives that
there exists $N\geq 1$ such that if $n>N$ then $|c_n|^{\frac{1}{n}}< \alpha+1$.
Then $|c_n|^{\frac{1}{n^2}}< (\alpha+ 1)^{\frac{1}{n}}$ for all $n>N$. 
It follows that $\limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n^2}}\leq 1$. 
Also, there is a subsequence $\{n_k\}$ such that $|c_{n_k}|^{\frac{1}{n_k}}$ converges to $\alpha$. For that subsequence, we have
$$
|c_{n_k}|^{\frac{1}{n_k^2}} \rightarrow 1
$$
as $k\rightarrow\infty$. 
This implies that $\limsup_{n\rightarrow\infty} |c_n|^{\frac{1}{n^2}}=1$.
Therefore, the radius of convergence of $\sum c_n x^{n^2}$ is 1. 
