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.

Express $|a_n|^{\frac1{n^2}}$ as $\left(|a_n|^{\frac1{n}}\right)^\frac1n$. Then show that $\limsup\left(|a_n|^{\frac1{n}}\right)^\frac1n=1.$

In general if $(b_n)_{n\in\mathbb N}$ is a positive sequence such that $\limsup b_n \in (0,+\infty)$ then $\limsup\left(b_n\right)^\frac1n=1.$
