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.

  • $\begingroup$ Ah, I had a feeling Cauchy-Shwarz would make an appearance. Thanks a lot! $\endgroup$ – yet_another_math_guy Jun 1 '16 at 4:31

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