Solution

L = lim sup_{n \to \infty} a_{n}^{1/ n}, R = \frac{1}{L} .

The 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 are

(a_{n}^{a_{n}})^{1/ n} = exp (\frac{a _{n} l o g a _{n}}{n}) .

Equality can again occur. For example, take

a_{n} = lo g (n + 2) .

Then

a_{n} \to \infty

a_{n}^{1/ n} \to 1

, and

\frac{a _{n} l o g a _{n}}{n} \to 0.

Thus both

\sum a_{n} z^{n}

and

\sum a_{n}^{a_{n}} z^{n}

Solved by gpt-5.6-sol high.

 Ancestors (11)