TITLE: Radius of convergence of $\sum\limits_{n=0}^\infty2^{-n^2}z^n$
QUESTION [6 upvotes]: I was reading examples to find the radius of convergence for  power series. The power series is defined as $\displaystyle\sum\limits_{n=0}^\infty c_n(z-z_0)^n$. And to find the radius of convergence $R$ we use $\displaystyle\limsup\limits_{n\rightarrow\infty} |c_n|^\frac{1}{n}=\frac{1}{R}$.

For $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}z^n$, how did they get that $\displaystyle\limsup\limits_{n\rightarrow\infty} |c_n|^\frac{1}{n} = \limsup\limits_{n\rightarrow\infty}(2^{-n^2})^\frac{1}{n}$? I can see they took $c_n = 2^{-n^2}$, but how does $c_n$ actually relate to $(2^{-n^2})$? And how does $(z-z_0)^n$ relate to $z^n$?
As in, im confused as how $\displaystyle\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ is related to $\displaystyle\sum\limits_{n=0}^\infty2^{-n^2}z^n$. What is $z^n$ and how is it linked to  $(z-z_0)^n$? What is $c_n$ and how is it related to $(2^{-n^2})$

REPLY [5 votes]: $z_0 =0$.  In general, $z_0$ is a fixed complex number that is the base point for the expansion.  $(z-0)^n=z^n$, so it sort of hides in this case.
$c_n=2^{-n^2}$, because that is the coefficient of $z^n=(z-0)^n$. Generally, $c_n$ is just whatever number is multiplied by $(z-z_0)^n$ in the power series.  So $c_n$ relates to $2^{-n^2}$ by being equal to it.