Power series are a fundamental tool in complex analysis, allowing us to represent functions as infinite sums of complex terms. They're like a supercharged version of polynomials, letting us describe complex functions with incredible precision within specific regions.

Understanding power series convergence is crucial. The radius of convergence tells us where a series behaves nicely, converging absolutely inside a disc and potentially diverging outside. This concept is key to working with and manipulating power series effectively.

Power series and convergence

Definition and properties of power series

• A power series is an infinite series of the form $\sum_{n=0}^{\infty} a_n(z-c)^n$, where $a_n$ are complex coefficients, $c$ is a complex number called the center of the series, and $z$ is a complex variable
• Power series can be used to represent complex functions in a specific domain around the center $c$
• The convergence of a power series depends on the values of $z$ and the behavior of the coefficients $a_n$

Radius and domain of convergence

• The radius of convergence $R$ is a non-negative real number or infinity that determines the largest open disc centered at $c$ in which the power series converges
  • Inside the disc of convergence ($|z-c| < R$), the power series converges absolutely
  • On the boundary of the disc ($|z-c| = R$), the series may converge conditionally, diverge, or converge absolutely
  • Outside the disc ($|z-c| > R$), the series diverges
• The radius of convergence can be determined using the ratio test or the root test on the coefficients $a_n$ of the power series
  • Ratio test: $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = L$, then $R = \frac{1}{L}$ (if $L = 0$, then $R = \infty$; if $L = \infty$, then $R = 0$)
  • Root test: $\lim_{n \to \infty} \sqrt[n]{|a_n|} = L$, then $R = \frac{1}{L}$ (if $L = 0$, then $R = \infty$; if $L = \infty$, then $R = 0$)

Domain of convergence

Finding the domain of convergence

• The domain of convergence is the set of all complex numbers $z$ for which the power series converges
• For a power series centered at $c$ with radius of convergence $R$, the domain of convergence is the open disc ${z \in \mathbb{C} : |z-c| < R}$
• To find the domain of convergence:
  1. Determine the radius of convergence $R$ using the ratio test or the root test on the coefficients $a_n$
  2. Investigate the behavior of the power series at the boundary points ($|z-c| = R$) to determine if they should be included in the domain of convergence
  3. Express the domain of convergence using set notation or interval notation, depending on whether the boundary points are included

Examples of finding the domain of convergence

• Example 1: For the power series $\sum_{n=0}^{\infty} \frac{(z-1)^n}{n!}$, the ratio test yields $R = \infty$, so the domain of convergence is $\mathbb{C}$ (the entire complex plane)
• Example 2: For the power series $\sum_{n=0}^{\infty} n!(z+2)^n$, the root test yields $R = \frac{1}{e}$, and the series converges at the boundary points. The domain of convergence is ${z \in \mathbb{C} : |z+2| \leq \frac{1}{e}}$ or $[-2-\frac{1}{e}, -2+\frac{1}{e}]$ in interval notation

Manipulating power series

Addition and multiplication of power series

• Power series can be added and multiplied term by term within their common domain of convergence
• To add two power series with the same center $c$, add their corresponding coefficients: $\sum_{n=0}^{\infty} a_n(z-c)^n + \sum_{n=0}^{\infty} b_n(z-c)^n = \sum_{n=0}^{\infty} (a_n+b_n)(z-c)^n$. The resulting series has the same center $c$ and a radius of convergence at least equal to the smaller of the two original radii
• To multiply two power series with the same center $c$, use the Cauchy product: $\left(\sum_{n=0}^{\infty} a_n(z-c)^n\right) \times \left(\sum_{n=0}^{\infty} b_n(z-c)^n\right) = \sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} a_k b_{n-k}\right)(z-c)^n$. The resulting series has the same center $c$ and a radius of convergence at least equal to the smaller of the two original radii

Division of power series

• To divide two power series with the same center $c$, use long division of power series
• The resulting series has the same center $c$ and a radius of convergence at least equal to the smaller of the two original radii, excluding any zeros of the denominator series
• Example: To divide $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$ by $1+z = \sum_{n=0}^{1} z^n$, perform long division to obtain $\frac{1}{1-z^2} = \sum_{n=0}^{\infty} (-1)^n z^{2n}$, with a radius of convergence $R = 1$

Differentiation and integration of power series

Term-by-term differentiation

• Power series can be differentiated term by term within their domain of convergence
• To differentiate a power series $\sum_{n=0}^{\infty} a_n(z-c)^n$ term by term, multiply each coefficient $a_n$ by $n$ and decrease the exponent of $(z-c)$ by 1: $\frac{d}{dz} \sum_{n=0}^{\infty} a_n(z-c)^n = \sum_{n=1}^{\infty} na_n(z-c)^{n-1}$. The resulting series has the same center $c$ and radius of convergence as the original series
• Example: Differentiating the geometric series $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$ yields $\frac{1}{(1-z)^2} = \sum_{n=1}^{\infty} nz^{n-1}$, with the same radius of convergence $R = 1$

Term-by-term integration

• Power series can be integrated term by term within their domain of convergence
• To integrate a power series $\sum_{n=0}^{\infty} a_n(z-c)^n$ term by term, divide each coefficient $a_n$ by $(n+1)$ and increase the exponent of $(z-c)$ by 1: $\int \left(\sum_{n=0}^{\infty} a_n(z-c)^n\right)dz = \sum_{n=0}^{\infty} \frac{a_n}{n+1}(z-c)^{n+1} + C$. The resulting series has the same center $c$ and radius of convergence as the original series
• Example: Integrating the geometric series $\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n$ yields $-\ln(1-z) = \sum_{n=0}^{\infty} \frac{z^{n+1}}{n+1} + C$, with the same radius of convergence $R = 1$