Question

Find the disk of convergence for each of the following complex power series. $$z-\frac{z^{2}}{2}+\frac{z^{3}}{3}-\frac{z^{4}}{4}+\cdots$$

Short answer

The disk of convergence is \(\left| z \right| < 1\).

Step-by-step solution

- Identify the general term

The given power series can be written in the general form. Identify the general term of the series. For the given series \[z - \frac{z^{2}}{2} + \frac{z^{3}}{3} - \frac{z^{4}}{4} + \cdots \], the general term can be expressed as: \[ a_n = \frac{(-1)^{n+1} z^n}{n} \]

- Use the Ratio Test for convergence

To find the radius of convergence, use the Ratio Test, which involves considering the limit of \(|a_{n+1}/a_n|\): \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| \] Simplify the expression inside the limit.

- Simplify the limit

Simplify the limit expression: \[\lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| = \lim_{n \to \infty} \left| \frac{z^{(n+1)}}{(n+1)} \cdot \frac{n}{z^n} \right| = \lim_{n \to \infty} \left| \frac{n \, z}{n+1} \right| = \left| z \right| \lim_{n \to \infty} \left| \frac{n}{n+1} \right| = \left| z \right| \]

- Determine the radius of convergence

The Ratio Test states that the series converges when this limit is less than 1: \[\left| z \right| < 1\]This inequality implies that the radius of convergence \(R\) is 1.

- Conclusion

Conclude that the disk of convergence is given by the set of \(z\) such that \(\left| z \right| < 1\).

Key concepts

Ratio Test

To determine the convergence of a complex power series, one useful method is the Ratio Test. This test involves taking the limit of the ratio of successive terms in the series. Specifically, for a series with terms \(a_n\), we calculate \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\.
The result of this limit helps us figure out whether the series converges or diverges:

• If the limit is less than 1, the series converges.
• If the limit is greater than 1, the series diverges.
• If the limit equals 1, the test is inconclusive.

In the specific problem of finding the disk of convergence for the series \(z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots\), the Ratio Test is applied to the terms of the series.

Radius of Convergence

The radius of convergence is a critical concept in understanding where a power series converges. This radius, denoted by \R\, defines a circle within which the series will converge.
Using the Ratio Test for the series \(z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots \), we find:
\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| \end{align*}
This cleans up to
\[ \left| z \right| < 1 \] This implies the radius of convergence \(R = 1\).

Disk of Convergence

The disk of convergence is the region in the complex plane where a power series converges. For the series given in the example, after finding the radius of convergence \(R = 1\), we determine that the series converges for all points inside the circle of radius 1 centered at the origin. This is written mathematically as:

• \left| z \right| < 1

Meaning, the series will converge for all \(z\) values that satisfy this condition. The term 'disk' comes from imagining a disk shape in the complex plane formed by these points.

Power Series

A power series is a type of series that takes the form:
\a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \cdots\
Here, each term is composed of constant coefficients \(a_n\) and powers of \(z\). The series provided in the problem statement can be rewritten in this general form:
\sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}\br>A complex power series converges when \(z\) is within a particular range, described by the radius of convergence. Outside of this range, the series may diverge.