Radius of Convergence from Limit of Sequence/Complex Case

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

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

Then the radius of convergence $R$ of $S \left({z}\right)$ is given by:


 * $(1) \quad \displaystyle \dfrac 1 R = \limsup_{n \to \infty} \left\vert{a_n}\right\vert^{1/n}$

If the sequence $\left \langle{\left\vert{\dfrac {a_{n+1}} {a_n}}\right\vert }\right\rangle_{n \in \N}$ converges, then $R$ is also given by:


 * $(2) \quad \displaystyle \dfrac 1 R = \lim_{n \to \infty} \left\vert{\dfrac {a_{n+1}} {a_n}}\right\vert$

If either


 * $\displaystyle \limsup_{n \to \infty} \left\vert{a_n}\right\vert^{1/n} = 0$

or


 * $\displaystyle \lim_{n \to \infty} \left\vert{\dfrac {a_{n+1}} {a_n}}\right\vert = 0$

then the radius of convergence is infinite, and $S \left({z}\right)$ is absolutely convergent.

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

Suppose that $\left\vert{z - \xi}\right\vert = R - \epsilon$.

By definition of radius of convergence, it follows that $S \left({z}\right)$ is absolutely convergent.

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


 * $\displaystyle \limsup_{n \to \infty} \left\vert{a_n} \left({z - \xi}\right)^n \right\vert^{1/n} \le 1$

By Combination Theorem for Sequences/Multiple Rule, this inequality can be rearranged to obtain:


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

As $\epsilon > 0$ was arbitrary, it follows that $\displaystyle \limsup_{n \to \infty} \left\vert{a_n}\right\vert^{1/n} \le \dfrac 1 R$.

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

Then $S \left({z}\right)$ is divergent, so the $n$th Root Test shows that:


 * $\displaystyle \limsup_{n \to \infty} \left\vert{a_n} \left({z - \xi}\right)^n \right\vert^{1/n} \ge 1$

which we can rearrange to obtain:


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

As $\epsilon > 0$ was arbitrary, it follows that $\displaystyle \limsup_{n \to \infty} \left\vert{a_n}\right\vert^{1/n} \ge \dfrac 1 R$.

It follows that $\displaystyle \limsup_{n \to \infty} \left\vert{a_n}\right\vert^{1/n} = \dfrac 1 R$, which is equation $(1)$.

Proof of $(2)$
Suppose that the sequence $\left\langle{ \left\vert{\dfrac {a_{n+1}} {a_n}}\right\vert }\right\rangle_{n \in \N}$ converges.

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

Suppose that $\left\vert{z - \xi}\right\vert = R - \epsilon$.

By definition of radius of convergence, it follows that $S \left({z}\right)$ is absolutely convergent.

Then, the Ratio Test shows that:


 * $\lim_{n \to \infty} \left\vert{\dfrac {a_{n+1} \left({z - \xi}\right)^{n+1} } {a_n \left({z - \xi}\right)^n} }\right\vert \le 1$

By Combination Theorem for Sequences/Multiple Rule, this inequality can be rearranged to obtain:

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

Then $S \left({z}\right)$ is divergent, so the Ratio Test shows that:


 * $\displaystyle \lim_{n \to \infty} \left\vert{\dfrac {a_{n+1} \left({z - \xi}\right)^{n+1} } {a_n \left({z - \xi}\right)^n} }\right\vert \ge 1$

Similarly, this inequality can be rearranged as:


 * $\displaystyle \lim_{n \to \infty} \left\vert{\dfrac {a_{n+1} } {a_n} }\right\vert \ge \dfrac{1}{R + \epsilon}$

As $\epsilon > 0$ was arbitrary, it follows that $\displaystyle \lim_{n \to \infty} \left\vert{\dfrac {a_{n+1}} {a_n}}\right\vert = \dfrac 1 R$, which is equation $(2)$.