Radius of Convergence from Limit of Sequence

Theorem

Real Case

Let $\xi \in \R$ be a real number.

Let $\ds S \paren x = \sum_{n \mathop = 0}^\infty a_n \paren {x - \xi}^n$ be a power series about $\xi$.

Then the radius of convergence $R$ of $S \paren x$ is given by:

$\ds \frac 1 R = \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} }$

if this limit exists and is nonzero.

If

$\ds \lim_{n \mathop \to \infty} \size {\frac {a_{n + 1} } {a_n} } = 0$

then the radius of convergence is infinite and therefore the interval of convergence is $\R$.

Complex Case

Let $\xi \in \C$ be a complex number.

Let $\ds S \paren z = \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$ be a (complex) power series about $\xi$.

Let the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converges.

Then $R$ is given by:

$\ds \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$

If:

$\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = 0$

then the radius of convergence is infinite, and $S \paren z$ is absolutely convergent for all $z \in \C$.