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.
Best Answer
The first can be done without recourse to $\limsup$:
If $R$ is the radius of convergence of $\sum_{n=0}^{\infty}c_n x^n$, then the series converges absolutely if $|x|<R$ and diverges if $|x| > R$.
Hence $\sum_{n=0}^{\infty}c_n x^{2n}= \sum_{n=0}^{\infty}c_n (x^2)^n$ converges absolutely if $|x^2| < R$, and diverges if $|x^2| >R$. Hence the radius of convergence must be $\sqrt{R}$.
For the second, note that if we write the power series as $\sum_{n=0}^{\infty}a_n x^n$, then $a_n = 0$ if $n$ is not a square, and $a_n =c_k$ if $n=k^2$. Hence the radius of convergence is $\frac{1}{\rho} = \limsup_n \sqrt[n^2]{|c_n|} $.
We have $\frac{1}{R} = \limsup_n \sqrt[n]{|c_n|}$, hence for all $\epsilon>0$, there exists a subsequence $n_k$ such that $\frac{1}{R} - \epsilon \le \sqrt[n_k]{|c_{n_k}|}$, and for all $\epsilon>0$, there exists $n$ such that $\frac{1}{R} + \epsilon \ge \sqrt[k]{|c_{k}|}$ for all $k \ge n$.
Hence we have $\sqrt[n_k]{\frac{1}{R} - \epsilon} \le \sqrt[n_k^2]{|c_{n_k}|}$, and taking $\limsup$s we have $1 \le \frac{1}{\rho}$. Similarly, the other inequality gives $1 \ge \frac{1}{\rho}$. Hence $\rho = 1$.