Radius of convergence of $\sum_{k=0}^{\infty} c_n^2x^n$. The problem statement is as follows:
Suppose the radius of convergence of the complex valued series $\sum_{n=0}^{\infty} c_nz^n$ is $R$.
Find the radius of convergence of $\sum_{n=0}^{\infty} c_n^2z^n$.
I've heard of a way to do this using the limit superior, however I'm wondering if there is a solution that doesn't involve this method.   I have been able to show that the radius of convergence is at least $R^2$:
for all $n$, we can show by induction that
$$
\sum_{n = 0}^N |c_n|^2|z|^{2n} \leq \left(\sum_{n = 0}^N |c_n||z|^n \right)^2. 
$$
Then if $w$ is such that $|w| < R^2$, then $\sqrt{|w|} < R$ and so
$$
\sum_{n = 0}^N |c_n|^2|w|^{n} \leq\left(\sum_{n = 0}^N |c_n|\sqrt{|w|}^n \right)^2. 
$$
As $\sqrt{|w|}$ lies in the open disk defined by radius of convergence of $\sum_{n=0}^{\infty} c_nz^n$, we have that this series converges for $z =\sqrt{|w|}$.
Thus by the comparison of limits, we have that the series $\sum_{n=0}^{\infty} c_n^2w^n$ converges absolutely for all such $|w|<R^2$.  Hence the radius of convergence must be greater than or equal to $R^2$.
The problem then becomes how do I show that the radius of convergence cannot be any larger than $R^2$.

If you take $z$ such that $|z|>R$, then choose $z'$ such that $|z| > |z'| > R$.  Then Cauchy-Schwartz gives
$$
\left(\sum_{i=0}^N |c_n|^2|z|^{2n}\right)
\left(\sum_{i=0}^N |z'/z|^{2n}\right)
\ge \left(\sum_{i=0}^N |c_n||z'|^{n}\right)^2
$$
This gives
$$
\sum_{i=0}^N |c_n|^2|z|^{2n} \ge \frac{|z|^2-|z'|^2}{|z|^2}\left(\sum_{i=0}^N |c_n||z'|^{n}\right)^2
$$
Choosing $w$ with $|z|^2 = |w|$ and $w'$ such that $|z'|^2 = |w'|$ gives
$$
\sum_{i=0}^N |c_n|^2|w|^{n} \ge \frac{|w|-|w'|}{|w|}\left(\sum_{i=0}^N |c_n|\left|\sqrt{w'}\right|^{n}\right)^2
$$
This shows the desired divergence.
