# Radius of convergence $\sum\limits_{n\ge0}q^{n^2}z^n$

Radius of convergence of the power series $\sum\limits_{n\ge0}q^{n^2}z^n$

By a theorem the radius is $\limsup \lvert q^{n^2}\rvert^{1/n}=\limsup\lvert q^n\rvert=\begin{cases}0,&\text{if}\ |q|<1\\1,&\text{if}\ |q|=1\\ \infty,&\text{if}\ |q|>1\end{cases}$

So either the series converges everywhere, for $|z|<1$, or nowhere, but for the middle case, is $|z|=1$ included or excluded ?

The correct formula is ${1\over R }= lim sup |q^{n^2}|^{1/n}$. So, the radius is $\infty$ for $|q|<1$ ( so, it converges everywhere ), the radius is $1$ for $|q|=1$ ( so, it converges in the open unit disc ) and the radius is $0$ for $|q|>1$ ( so, it diverges everywhere ). The circle $|z|=1$ is not to be included because there are some points on it where the series converges and on other points it diverges. For example, take the point $z=1$. we will get the sum to be divergent. It cannot even be analytically continued beyond the open unit disc by hadamard's gap theorem.