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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 $(2^{-n^2})$

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    Thanks @JonasMeyer, maybe it was not explained in my notes, but what exactly is $c_n$ and how is that linked to $2^{-n^2}$?2012-05-24

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$z_0 =0$. 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.

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    ...then $f(z)=\sum\limits_{n=0}^\infty \frac{(-1)^{n+1}}{2^{n+1}}(z-3)^n$, with radius of convergence $2$.2012-05-24