# Power Series¶

A series of the form of
$$f(z) = \sum_{n=0}^\infty a_n (z-a)^n \quad a_n \in \mathbb{C}$$ is called a power series. Here, $$a$$ is called the center of the power series.

Definition

For any power series there exist a real number $$R\geq 0$$, such that, the given power series converges in open disc of radius $$R$$, i.e., $$\vert z - a \vert < R$$, while diverges for all $$z$$ outside the closed disc of radius $$R$$, i.e., $$\vert z - a \vert > R$$. Such $$R$$ is called the radius of convergence the given power series.

Warning

At the circle of convergence, i.e., at $$\vert z - a\vert = R$$, the power series can converge for some $$z$$ or diverge for some other points. For example consider the power series, $$f(z) = \sum_{n=0}^\infty \dfrac{z^n}{n},$$ the radius of convergence will be $$1$$. It diverge for $$z=1$$ but converge for $$z=-1$$, both lie on the circle of convergence.

## 2. Method of finding Radius of Convergence¶

Two theorem from real analysis usualy used to find the radius of convergence as mentioned below.

Ratio Test

For the power series $$\sum_{n=0}^\infty a_n (z-a)^n$$, consider the followin limit exist, $$L = \lim_{n\to \infty}\left \vert \dfrac{a_n}{a_{n+1}}\right\vert.$$ Then

1. If $$L < 1$$, then the series absolutely converges.
2. If $$L > 1$$, then the series diverges.
3. If $$L = 1$$, then the series is either divergent or convergen
Hence, the radius of convergence will be $$R = 1/L$$.

For the power series $$\sum_{n=0}^\infty a_n (z-a)^n$$, the radius of converge is given by $$\(\dfrac{1}{R} = \limsup_{n\to \infty} \left\vert a_n \right \vert ^{\dfrac{1}{n}}$$\)