Radius of Convergence from Limit of Sequence/Complex Case

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

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

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


 * $(1): \quad \displaystyle \dfrac 1 R = \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n}$

If the sequence $\sequence {\cmod {\dfrac {a_{n + 1} } {a_n} } }_{n \mathop \in \N}$ converges, then $R$ is also given by:


 * $(2): \quad \displaystyle \dfrac 1 R = \lim_{n \mathop \to \infty} \cmod {\dfrac {a_{n + 1} } {a_n} }$

If either


 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = 0$

or


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

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

Proof of $(1)$
Let $\epsilon \in \R_{>0}$, and let $z \in \C$.

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

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

Then, the $n$th Root Test shows that:


 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \le 1$

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


 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 {\cmod {z - \xi} } = \dfrac 1 {R - \epsilon}$

As $\epsilon > 0$ was arbitrary, it follows that:
 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \le \dfrac 1 R$

Now, suppose that $\cmod {z - \xi} = R + \epsilon$.

Then $S \paren z$ is divergent, so the $n$th Root Test shows that:


 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n \paren {z - \xi}^n}^{1/n} \ge 1$

which we can rearrange to obtain:


 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 {\cmod {z - \xi} } = \dfrac{1}{R - \epsilon}$

As $\epsilon > 0$ was arbitrary, it follows that:
 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} \ge \dfrac 1 R$

It follows that:
 * $\displaystyle \limsup_{n \mathop \to \infty} \cmod {a_n}^{1/n} = \dfrac 1 R$

which is equation $(1)$.

Proof of $(2)$
Suppose that the sequence $\sequence {\cmod {\dfrac {a_{n+1}} {a_n} } }_{n \mathop \in \N}$ converges.

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

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

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

Then, the Ratio Test shows that:


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

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

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

Now, suppose that $\cmod {z - \xi} = R + \epsilon$.

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


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

Similarly, this inequality can be rearranged as:


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

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

which is equation $(2)$.