A sequence tending to zero

Convergence is the easy part

Let c > 0, u₀ = c and u(n+1) = ln(1 + u(n)) for n ≥ 0.

Two remarks suffice:

• if u(n) > 0, then 1 + u(n) > 1, so u(n+1) = ln(1 + u(n)) > 0. By induction the sequence stays strictly positive, and the iteration is always defined;
• the inequality ln(1 + x) ≤ x from part I, applied to x = u(n), gives u(n+1) ≤ u(n): the sequence is decreasing.

Decreasing and bounded below by 0, it converges to a real number ℓ ≥ 0. Since x ↦ ln(1 + x) is continuous, the limit satisfies ℓ = ln(1 + ℓ). But ln(1 + x) < x as soon as x > 0, hence ℓ = 0.

The real question: how fast?

Knowing that u(n) tends to 0 says nothing about the speed. A sequence may tend to 0 like 1/n, like 1/√n or like 2⁻ⁿ, and these three behaviours have nothing in common.

The assignment answers by studying the product n·u(n). If it converges to a finite non-zero limit, then u(n) behaves exactly like that limit divided by n: we know the speed, not merely the order of magnitude.