Question

Find the disk of convergence for each of the following complex power series. $$1-\frac{z^{2}}{3 !}+\frac{z^{4}}{5 !}-\cdots$$

Short answer

The radius of convergence is infinite.

Step-by-step solution

- Recognize the General Form of the Series

Identify the general term of the power series. The given series is: i.e., $$1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$$Notice that the series can be written in terms of a general term as:$$a_n z^{2n} = \frac{(-1)^n z^{2n}}{(2n+1)!}$$.

- Apply the Ratio Test

To find the radius of convergence, use the Ratio Test on the general term:$$a_n = \frac{(-1)^n}{(2n+1)!}$$Compute the limit of the absolute value of the ratio of successive terms:$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{(2(n+1)+1)!}}{\frac{(-1)^n}{(2n+1)!}} \right|$$where we simplify the absolute value as:$$\lim_{n \to \infty} \left| \frac{(2n+1)!}{(2(n+1)+1)!} \right| = \lim_{n \to \infty} \left| \frac{(2n+1)!}{(2n+3)!} \right|$$.

- Simplify the Ratio

Simplify the expression inside the limit:$$\lim_{n \to \infty} \left| \frac{(2n+1)!}{(2n+3)(2n+2)(2n+1)!} \right| = \lim_{n \to \infty} \left| \frac{1}{(2n+3)(2n+2)} \right|$$This further simplifies to:$$\lim_{n \to \infty} \left| \frac{1}{(2n+3)(2n+2)} \right| = 0$$.

- Find the Radius of Convergence

Based on the Ratio Test, convergence occurs when the result is less than 1. Since the limit $$\lim_{n \to \infty} \left| \frac{1}{(2n+3)(2n+2)} \right| = 0$$ is always less than 1 for all values of \(z\), the series converges for all \(z\). Hence the radius of convergence is infinite.

Key concepts

Complex Power Series

A complex power series is a series of the form \( \sum_{n=0}^{\infty} a_n z^n \), where \(a_n\) are coefficients and \(z\) is a complex number. These series generalize power series to the complex plane.
In the given exercise, the series is represented with even powers of \(z\) and factorial denominators, given as: $$1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$$.
Understanding these series is important because they can represent complex functions in terms of simpler components.

• Start by identifying the general term. For this series, we can write it as \( a_n z^{2n} \).
  
• Recognize the pattern to derive the general term for any value of \(n\).

For this problem, we identify it as:
$$ a_n z^{2n} = \frac{(-1)^n z^{2n}}{(2n+1)!} $$.
Understanding this form helps use further methods to determine convergence, such as the Ratio Test.

Ratio Test

The Ratio Test is a common technique to determine the convergence of a series. It involves computing the limit of the ratio of successive terms. For a series \( \sum a_n \), you evaluate:
\( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
If this limit is less than 1, the series converges.
In our exercise, we applied the Ratio Test to:
$$ a_n = \frac{(-1)^n}{(2n+1)!} $$ to form the ratio:
$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{\frac{(-1)^{n+1}}{(2(n+1)+1)!}}{\frac{(-1)^n}{(2n+1)!}} \right|. $$

• Factorials grow very fast, so simplifying this leads us to explore the rapid descent of the terms.

Further simplifying gives:
$$ \lim_{n \to \infty} \left| \frac{(2n+1)!}{(2(n+1)+1)!} \right| = \lim_{n \to \infty} \left| \frac{(2n+1)!}{(2n+3)!} \right|. $$
Continuing this process, and simplifying ultimately to:
$$ \lim_{n \to \infty} \left| \frac{1}{(2n+3)(2n+2)} \right|, $$
which evaluates to 0. Since 0 is less than 1, this confirms convergence for all \(z\).

Infinite Radius of Convergence

The radius of convergence describes the range of values around the center of the series for which the series converges.
The radius can be found using results from the Ratio Test, which in the exercise gave:
$$ \lim_{n \to \infty} \left| \frac{1}{(2n+3)(2n+2)} \right| = 0. $$
Since the limit is 0, it implies convergence for any complex number \(z\).

• If the limit is zero, the series converges everywhere in the complex plane.

Hence, for this series, the radius of convergence is said to be infinite.
An infinite radius of convergence means the series converges for all \(z\) in the complex plane.
This is a significant characteristic, often found in series that have rapidly decreasing terms in the denominator, like factorials. Understanding this can simplify the analysis of complex functions as power series.