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The complex geometric power series has radius $1$ and sums to $1 / (1 - z)$ for $∣ z ∣ < 1$

Example

For $∣ z ∣ < 1$ , the complex geometric series $\sum_{n \geq 0} z^{n}$ has radius $1$ and sum $1 / (1 - z)$ . The conventions and prerequisite facts used below are recorded in Cauchy-Hadamard for complex power series, including zero and infinite radius, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ , Every absolutely convergent complex series converges, and rearrangements preserve its sum, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions.

Facts & Assumptions

Given: A complex $z$ with $∣ z ∣ < 1$ .

[L1]

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges says that the real series $\sum_{n \geq 0} r^{n}$ converges for $∣ r ∣ < 1$ .

[L2]

Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive gives $∣ z^{n} ∣ = ∣ z ∣^{n}$ .

[L3]

Every absolutely convergent complex series converges, and rearrangements preserve its sum says that an absolutely convergent complex series converges.

[L4]

For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ gives $∣ z ∣^{n} \to 0$ when $∣ z ∣ < 1$ .

[L5]

Cauchy-Hadamard for complex power series, including zero and infinite radius defines the shifted coefficient limsup and its radius cases.

Verification

1.1

The shifted coefficient roots in [L5] are all $1$ , so it gives radius $1$ .

2.1

By [L2], the modulus series is the real geometric series $\sum ∣ z ∣^{n}$ , which converges by [L1]; therefore the complex series converges by [L3], say to $S$ . The finite identity $(1 - z) \sum_{k < n} z^{k} = 1 - z^{n}$ holds in the complex field. Since $∣ z^{n} ∣ = ∣ z ∣^{n} \to 0$ by [L2] and [L4], passing to the limit gives $(1 - z) S = 1$ ; since $z \neq 1$ , $S = 1 / (1 - z)$ .

L1L2L3L4∎

Depends on

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Sources

J. Lebl, Basic Analysis I: Complex Numbers and the Complex Exponential (standard reference, not scraped)

The complex geometric power series has radius 11 and sums to 1/(1−z)1/(1−z) for ∣z∣<1∣z∣<1