 Solution

By the Cauchy-Hadamard theorem, putThe squared coefficients have root limsup $L^2$, so their radius is $R^2$. Equal finite nonzero radii therefore require $R=R^2$, hence $R=1$. This can happen: for $a_n=n+1$, both $\sum a_n{z}^n$ and $\sum a_n^2{z}^n$ have radius one.

For the coefficients $a_n^{a_n}$, the relevant root terms areEquality can again occur. For example, takeThen $a_n\to \infty$, $a_n^{1/n}\to 1$, andThus both $\sum a_n{z}^n$ and $\sum a_n^{a_n}{z}^n$ have radius one. These examples illustrate the radius of convergence after powering coefficients.