Absolute and Conditional Convergence; Rearrangement; Products: Examples and Counterexamples

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1 · Prerequisites

Real Analysis

Set Theory

2 · Logical flowchart

3 · Items

∑_j ≥ 0 (-1)ʲ/(j+1) converges conditionally, with sum strictly between 1/2 and 1

Every rearrangement of ∑_k ≥ 0 (-1/2)ᵏ converges to 2/3

Taking two positive terms for each negative one rearranges the alternating harmonic series to 3/2 times its sum, by the identity T₃ₙ = S₄ₙ + 1/2 S₂ₙ

An explicit greedy rearrangement of the alternating harmonic series with sum 0, and the same recipe for any prescribed real

The period-three pattern 1, 1, -2 has partial sums in {0,1,2}, so ∑ aₖ/(k+1) converges by Dirichlet's test although the alternating series test does not apply

∑_j ≥ 0 (-1)ʲ (j+3)/(j+1)² converges, by Abel's test with the monotone bounded factor (j+3)/(j+1)

With aⱼ = (-1)ʲ/√j+1 convergent and bⱼ = (-1)ʲ bounded but not monotone, ∑ aⱼ bⱼ = ∑ 1/√j+1 diverges

For |r| < 1 the Cauchy product of ∑ rᵏ with itself is ∑ (k+1) rᵏ, with sum 1/(1-r)²

The Cauchy product of ∑_k ≥ 0 (-1)ᵏ/√k+1 with itself has |cₙ| ≥ 1 for every n, so it diverges

The array with aᵢᵢ = 1, a_i+1,i = -1 and every other entry 0 has iterated sums 1 and 0

(1-1) + (1-1) + … converges to 0 while ∑ₖ (-1)ᵏ diverges

∏_j ≥ 0 (1 - 1/(j+2)) has partial products 1/(n+1), which tend to 0, so the product does not converge in the sense used here

∏_j ≥ 0 (1 + (-1)ʲ/√j+2) has partial products tending to 0 although ∑_j ≥ 0 (-1)ʲ/√j+2 converges

0.999… = 1 and 0.4999… = 0.5: the second expansion of a number is exactly an eventually-all-(b-1) digit sequence

Sources