ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-31

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The geometric series has only one singular point on its unit circle

Example

The geometric series

$\sum_{n \geq 0} z^{n} = \frac{1}{1 - z} (∣ z ∣ < 1)$

has radius $1$ . Its only singular boundary point on the unit circle is $z = 1$ . Every other boundary point is regular because the rational function $1 / (1 - z)$ is holomorphic there.

Facts & Assumptions

Given: The geometric series $\sum_{n \geq 0} z^{n}$ .

[L1]

For $∣ z ∣ < 1$ , the real geometric series $\sum_{n \geq 0} ∣ z ∣^{n}$ converges, so $\sum_{n \geq 0} z^{n}$ converges absolutely; the finite identity $(1 - z) \sum_{k < n} z^{k} = 1 - z^{n}$ holds in the complex field, and $∣ z^{n} ∣ = ∣ z ∣^{n} \to 0$ . (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Every absolutely convergent complex series converges, and rearrangements preserve its sum, $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ )

[L2]

The coefficient sequence $1, 1, 1, \dots$ gives radius $1$ by Cauchy-Hadamard (Cauchy-Hadamard for complex power series, including zero and infinite radius).

Verification

technique · direct

1.1L1L2

For $∣ z ∣ < 1$ , fact [L1] gives absolute convergence of $\sum_{n \geq 0} z^{n}$ , the finite identity $(1 - z) \sum_{k < n} z^{k} = 1 - z^{n}$ , and the limit $z^{n} \to 0$ . Passing to the limit yields the sum formula $\sum_{n \geq 0} z^{n} = 1 / (1 - z)$ on the unit disc, and [L2] gives radius $1$ .

2.1step 1.1algebra∎

If $ζ$ lies on the unit circle and $ζ \neq 1$ , then $1 - ζ \neq 0$ , so the same rational function is holomorphic on a neighbourhood of $ζ$ . Thus $ζ$ is regular. At $z = 1$ the denominator vanishes, so no holomorphic extension through $1$ can equal $1 / (1 - z)$ on a punctured neighbourhood. Therefore $1$ is the unique singular boundary point.

Depends on

Used by

FALSE: every boundary point of a radius-one power series is singular False statement

Dependency tree · two levels

45 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Curtis T. McMullen, Riemann Surfaces, Ch. 4 Example 2 (standard reference, not scraped)
Lars V. Ahlfors, Complex Analysis, 3rd ed., Ch. 2 (standard reference, not scraped)