Radius of convergence

Definition

Over the complex numbers

Consider the power series about a point $z_0\in \mathbb{C}$ with coefficients $a_n\in \mathbb{C}$:

$\sum a_n(z-z_0{)}^n$

The radius of convergence of this power series is defined in the following equivalent ways:

• It is defined as the largest $R$ such that the power series converges absolutely for all $z$ with $|z-z_0|<R$ and diverges for all $|z-z_0|>R$.
• It is given by the formula:

$R=\frac{1}{limsup{\left|a_n\right|}^{1/n}}$

If the denominator is $\mathrm{\infty }$, the radius of convergence is defined as 0, and if the denominator is 0, the radius of convergence is taken to be $\mathrm{\infty }$.

The open disk centered at $z_0$ and of radius equal to $R$ is termed the disk of convergence.

Over the real numbers

Consider the power series about a point $x_0\in \mathbb{R}$ with coefficients $a_n\in \mathbb{R}$:

$\sum a_n(x-x_0{)}^n$

The radius of convergence of this power series is defined in the following equivalent ways:

• It is defined as the largest $R$ such that the power series converges absolutely for all $x$ with $|x-x_0|<R$ and diverges for all $|x-x_0|>R$.
• It is given by the formula:

$R=\frac{1}{limsup{\left|a_n\right|}^{1/n}}$

If the denominator is $\mathrm{\infty }$, the radius of convergence is defined as 0, and if the denominator is 0, the radius of convergence is taken to be $\mathrm{\infty }$.

Facts

• Power series is infinitely differentiable in disk of convergence: This result holds both over the reals and the complex numbers.