Radius of Convergence from Limit of Sequence/Complex Case

Theorem
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 $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converge.

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 $z \in \C$.

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

Let $\epsilon \in \R_{>0}$, and let $z \in \C$.

Let $\cmod {z - \xi} = R - \epsilon$.

By definition of radius of convergence, it follows that $S \paren z$ is absolutely convergent.

Then from the Ratio Test:


 * $\lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n+1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \le 1$

By Multiple Rule for Complex Sequences, this inequality can be rearranged to obtain:

{{eqn | r = \dfrac 1 {\cmod {x - \xi} }}

Let $\cmod {z - \xi} = R + \epsilon$.

Then $\map S z$ is divergent, so the Ratio Test shows that:


 * $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} \paren {z - \xi}^{n + 1} } {a_n \paren {z - \xi}^n} } \ge 1$

Similarly, this inequality can be rearranged as:


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

As $\epsilon > 0$ was arbitrary, it follows that:
 * $\ds \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} } = \dfrac 1 R$

Hence the result.

Also presented as
This result can also be seen presented as:


 * $\ds R = \lim_{n \mathop \to \infty} \cmod {\frac {a_{n - 1} } {a_n} }$

but in this case the condition under which the radius of convergence is infinite is less conveniently stated.