ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-16

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The harmonic complex power series diverges at 1 and converges conditionally at every other point of the unit circle

Example

For $∣ z ∣ = 1$ , the series $\sum_{n = 1}^{\infty} z^{n} / n$ diverges at $z = 1$ and converges conditionally at every $z \neq 1$ .

Facts & Assumptions

Given: A complex number $z$ with $∣ z ∣ = 1$ .

[L1]

Abel summation gives the finite summation-by-parts and tail identities for complex coefficients (Abel summation by parts for complex coefficients and their partial sums).

[L2]

For real $p$ , the real series $\sum_{n \geq 1} 1 / n^{p}$ converges exactly when $p > 1$ (The p-series for a real exponent p converges exactly when p is greater than one).

[L3]

Given a positive real $ε$ , there is a natural number $N \geq 1$ with $1 / N < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L4]

Complex modulus is multiplicative and satisfies the triangle inequality (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Verification

technique · direct

1.1L4algebra

If $z \neq 1$ , the finite identity $(1 - z) \sum_{k = 1}^{N} z^{k} = z - z^{N + 1}$ gives $∣ \sum_{k = 1}^{N} z^{k} ∣ \leq 2 / ∣ 1 - z ∣$ by [L4].

2.1step 1.1L1L3

Apply [L1] to the bounded partial sums in step 1.1 and the decreasing weights $1 / n$ . The tail is bounded by a fixed multiple of $1 / p$ , which tends to $0$ by [L3], so the series converges.

3.1step 2.1L2L4∎

At $z = 1$ the series is the divergent $p$ -series with $p = 1$ by [L2]. At every other point on the circle, $∣ z^{n} / n ∣ = 1 / n$ , so the modulus series also diverges by [L2]; the convergence from step 2.1 is therefore conditional. No term $1 / 0$ is formed.

Depends on

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